In-Class Problems 2 and 3: Projectile Motion Solutions. In-Class Problem 2: Throwing a Stone Down a Hill

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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Deprtment of Physics Physics 8T Fll Term 4 In-Clss Problems nd 3: Projectile Motion Solutions We would like ech group to pply the problem solving strtegy with the four stges (see below) to nswer the following two problems I Understnd get conceptul grsp of the problem II Devise Pln - set up procedure to obtin the desired solution III Crry our your pln solve the problem! IV Look Bck check your solution nd method of solution For the first problem we hve posed series of questions for ech of the bove steps to help you lern how to use the problem solving strtegy We then leve spce for your group to nswer the question You don t need to nswer ll these questions but they should help you pproch the problem In the second problem, we invite you to try to solve it without help If you would like some hints, we do pose series of hints to consider In-Clss Problem : Throwing Stone Down Hill A person is stnding on top of hill which slopes downwrds uniformly t n ngle φ with respect to the horizontl The person throws stone t n initil ngle θ from the horizontl with n initil speed of v You my neglect ir resistnce How fr below the top of the hill does the stone strike the ground?

2 I Understnd get conceptul grsp of the problem How would you model the horizontl nd verticl motions of the stone? Drw grph nd coordinte system Where did you choose your origin? Wht choices did you mke for positive xes? Do your choices for positive directions ffect the signs for position, velocity, or ccelertion? In terms of your coordinte system, is there ny constrint condition for the horizontl nd verticl position of the stone when it hits the ground in terms of the specified quntities in the problem (be creful with signs)? Answer: The problem involves constnt velocity in the horizontlly direction nd constnt ccelertion in the verticl direction y I will choose Crtesin coordinte system with the origin t the point the stone is thrown with verticl upwrds positive nd horizontl to the right s positive Therefore y = g The key constrint condition is tht the point where the stone strikes the hill hs y f < nd x f > Therefore tn φ = yf / x f for φ < or tn φ = yf / x f for φ > I will choose φ > II Devise Pln - set up procedure to obtin the desired solution In terms of your choices tht you mde in your model, wht equtions re pplicble to this problem? Identify the different symbols tht pper in your eqution Are they treted s knowns or unknowns? Do you hve enough equtions to solve the problem? Identify the quntity tht you would like to solve for nd design strtegy to find it Pln: I write down the horizontl nd verticl equtions for projectile motion x () t = x + v x, t yt () = y + v y, t gt

3 with the following initil nd finl conditions x =, v x, = v cosθ, y =, nd v y, = v sinθ The finl conditions re yt ( f )= y f nd x ( t f )= x f Thus the equtions of motion become x f = v cos θ t f y = v sinθ t f gt f f nd the constrint x f = y /tn φ f III Crry our your pln solve the problem! Be creful with signs!!!!! I m trying to solve for the finl verticl position y f I will solve the horizontl eqution for t f yielding t f = x f / v cos θ t f I will introduce the constrin condition t f = y f / v cos θ tn φ I will substitute this vlue for t f into the verticl eqution y = v sin θ t f gt f = v sin θ ( y / v cos θ tn φ) f f g y / v cos θ tn φ) ( f This eqution cn be clened up = tn θ / tn φ gy / v f cos θ tn φ Now I cn solve this for y f y = v cos θ tn φ f (+ tn θ / tn φ) g 3

4 IV Look Bck check your solution nd method of solution Does your solution mke sense? Check specil cses: wht do you expect for n nswer if the person is throwing the stone on level ground, or throwing the stone over verticl cliff Does your solution gree in these limits? Answer: If φ = I ned to be creful becuse this implies tht tn φ = so I cnnot divide by tn φ = Insted I need to note tht y f = Then the equtions of motion become x f = v cos θ t f = v sin θ t f gt f I cn solve the second eqution for t f, t f = v sin θ / g nd then for x f x f = v cos θ sin θ / g = v in( θ ) / g If φ = π / then tn π / = nd v cos θ tn φ y = v cos θ tn φ f ( + tn θ / tn φ) = = g g The stone never hits the ground 4

5 In-Clss Problem 3: Hitting the Bucket A person is stnding on ldder holding pil The person releses the pil from rest t height h bove the ground A second person stnding horizontl distnce s from the pil ims nd throws bll the instnt the pil is relesed in order to hit the pil The person throws the bll t height h bove the ground, with n initil speed v, nd t n ngle θ with respect to the horizontl You my ignore ir resistnce Questions: ) Find n expression for the ngle θ tht the person ims the bll in order to hit the pil s function of the other vribles given in the problem b) If the person ims correctly, find n expression for the rnge of speeds tht the bll must be thrown t in order to hit the pil? In cse you need some help, try nswering these questions I Understnd get conceptul grsp of the problem Sketch the motion of ll the bodies in this problem Introduce coordinte system Sketch nd Coordinte system: Sketch of motion: 5

6 c) Find n expression for the ngle θ tht the person throws the bll s function of h, h, nd s d) Find n expression for the time of collision s function of the initil speed of the bll v, nd the quntities h, h, nd s e) Find n expression for the height bove the ground where the collision occurred s function of the initil speed of the bll v, nd the quntities h, h, nd s f) Find n expression for the rnge of speeds (s function of h, h, nd s ) tht the bll cn be thrown in order tht the bll will collide with the pil? There re two objects involved in this problem Ech object is undergoing free fll, so there is only one stge ech The pil is undergoing one dimensionl motion The bll is undergoing two dimensionl motion The prmeters h, h, nd s re unspecified, so our nswers will be functions of those symbolic expressions for the quntities Since the ccelertion is unidirectionl nd constnt, we will choose Crtesin coordintes, with one xis long the direction of ccelertion Choose the origin on the ground directly underneth the point where the bll is relesed We choose upwrds for the positive y-direction nd towrds the pil for the positive x-direction We choose position coordintes for the pil s follows The horizontl coordinte is constnt nd given by x = s The verticl coordinte represents the height bove the ground nd is denoted by y () t The bll hs coordintes ( x (t), y ( t)) We show these coordintes in the figure below 6

7 II Devise Pln - set up procedure to obtin the desired solution Question: Wht equtions of motion follow from your model for the position nd velocity functions of ech body? Model: The pil undergoes constnt ccelertion ( y ) = g in the verticl direction downwrds nd the bll undergoes uniform motion in the horizontl direction nd constnt ccelertion downwrds in the verticl direction, with ( x ) = nd ( y ) = g Equtions of Motion for Pil: The initil conditions for the pil re (v ) =, x = s, ( y = h y, ) verticlly, the pil lwys stisfies the constrint condition x = s equtions for position nd velocity of the pil simplify to Equtions of Motion for Bll: y t h gt ()= v y, () t = gt Since the pil moves nd v x, = The The initil position is given by ( x ) =, ( y ) = h The components of the initil velocity re given by (v y, ) = v sin (θ ) nd (v x, ) = v cos (θ ), where v is the mgnitude of the initil velocity nd θ is the initil ngle with respect to the horizontl So the equtions for position nd velocity of the bll simplify to x (t) = v cos (θ )t v () t = v cos (θ ) x, y ( t )= h + v sin (θ )t gt v () t = v sin y, (θ ) gt 7

8 Question: How mny independent equtions nd unknowns do you hve? Should the quntities h, h, nd s be treted s knowns or unknowns Answer: Note tht the quntities h, h, nd s should be treted s known quntities lthough no numericl vlues were given, only symbolic expressions There re six independent equtions with 9 s yet unspecified quntities y ( t), t, y (), t x (t ), v (), t v (), t v (), t v, θ y, y, x, y t h gt ()= v y, () t = gt x (t) = v cos (θ )t v () t = v cos (θ ) x, y ( t )= h + v sin (θ )t gt v () t = v sin y, (θ ) gt So we need two more conditions, in order to find expressions for the initil ngle, θ, the time of collision, t, nd the sptil loction of the collision point specified by y t ) or y t in terms of the one unspecified prmeter v ( ) Question: Wht mthemticl formule follow from the phse hits the pil? Answer: At the collision time t = t, the collision occurs when the two blls re locted t the sme position Therefore ( )= ( ), nd x (t ) = x = s y t y t Question: Clen up your equtions Wht strtegy cn you design for finding the ngle the second person needs to im the bll? Answer: We shll pply the conditions we found for the bll hitting the pil ( 8

9 = + θ h gt h v sin ( )t gt s = v cos (θ )t From the first eqution, the term gt cncels from both sides Therefore we hve tht h = h + v sin (θ )t s = v cos (θ )t We will now solve these equtions for tn( θ ) = sin( θ ) / cos( θ ), nd thus the ngle the person throws the bll in order to hit the pil III Crry our your pln solve the problem! We rewrite these equtions s v sin (θ )t = h h Dividing these equtions yields v cos (θ )t = s v sin (θ )t v cos ( )t θ θ s = tn ( )= h h So the initil ngle is independent of v, nd is given by θ = tn (( h h ) s ) h h From the figure below we cn see tht tn ( )=, implies tht the second person θ s ims the bll t the initil position of the pil 9

10 Question: When does the bll collide with the pil? Answer: In order to find the time of collision s function of the initil speed, we begin with our results tht v sin (θ )t h h = v cos (θ )t = s We squre both of the equtions bove nd utilize the trigonometric identity So our squred equtions become sin (θ ) + cos (θ ) = v sin ( θ )t = (h h ) v cos (θ )t = s Adding these equtions together yields v sin ( θ )t + v cos ( θ )t = v t = s + ( h h ) We cn solve this for the time of collision s h h ) + ( t = v Question: At wht height does the bll collide with the pil? Answer: We cn use the y-coordinte function of either the bll or the pil t t = t Since it hd no initil y velocity, it s esier to use the pil,

11 g s h h ( )= ( + ( y t h v ) ) Question: Wht is the minimum speed the person must throw the bll in order to ensure tht there will be collision? Answer: Suppose the bll nd the pil collide exctly t the ground t the time t = t b The condition is tht the speed v must be gret enough such tht the bll reches x t s ( b )= before the bll hits the ground, y t y(t b ) = So we require tht ( ) = b x (t b ) = v cos (θ )t b s This condition is esiest to pply when solving for the time, t = t b, tht the pil hits the ground, y t h gt ( )= b b = Thus Therefore the condition becomes h t b = g t v cos (θ ) h x ( b )= s g or v g s = g s h cos ( θ ) h cos (tn ((h h ) s ))

12 IV Look Bck check your solution nd method of solution Question: check your lgebr nd your units Any obvious errors? Answer: No obvious errors Question: Cn you see tht the nswer is correct now tht you hve it often simply by retrospective inspection? Answer: The person ims t the pil t the point where the pil ws relesed Both undergo free fll so the key result ws tht the verticl position obeys = + θ h gt h v sin ( )t gt The distnce trveled due to grvittionl ccelertion re the sme for both so ll tht mtters is the contribution form the initil positions nd the verticl component of velocity h = h + v sin (θ )t Since the time is relted to the horizontl distnce by s = v cos (θ )t This is now s if both objects were moving t constnt velocity Question: Substitute some vlues for the initil conditions tht test the limits of your nswer Mke the pil very fr wy or very close nd wht your nswer predicts bout the time of flight or the collision height Answer: Let s = m, v = m s -, let h = m, nd h = 4m Then the condition for the initil speed is stisfied since v s h cos (tn ((h h s g ) m s - m = = 4 m s- )) (4 m) cos (tn ((4 m m ) m )) Question: Cn you solve it different wy? Is the problem equivlent to one you ve solved before if the vribles hve some specific vlues?

13 Answer: This is n unusul ppliction of moving to reference frme ccelerting downwrds with A = g Then the problem is simply y y t h ()= x t v cos (θ )t ()= y ()= t h + v sin (θ )t 3