PHYS Summer Professor Caillault Homework Solutions

Transcription

1 PHYS Summe Pofesso Caillault Homewok Solutions Chapte 4 3. Pictue the Poblem: The ca moves up the 5.5 incline with constant acceleation, changing both its hoizontal and vetical displacement simultaneously. Stategy: Find the magnitude of the displacement along the incline, and then independently find the hoizontal and vetical components of the displacement. Solution: 1. (a) Find the magnitude of the displacement along the incline using equation -11: Δ 0 + t + 1 at (.0 m/s ) 1 s. Find the hoizontal component of Δ : x d cosθ ( 140 m)cos m 3. (b) Find the vetical component Δ : y d sinθ ( 140 m)sin m ( ) 140 m Insight: The hoizontal and vetical motions can be consideed sepaately. In this case they ae each descibed by constant acceleation motion, but the vetical acceleation is less than the hoizontal. The two acceleations would be equal if the angle of the incline wee Pictue the Poblem: The baseball is launched hoizontally and follows a paabolic tajectoy like that pictued at ight. Stategy: Using equations 4-7 find the time it takes the baseball to tavel the 18 m to the plate, and then calculate the vetical distance it will fall duing that time. Solution: 1. (a) Find the time it takes the ball to each the plate knowing that its hoizontal velocity emains unchanged thoughout the flight: t x 18 m 3 m/s s ( ) s. Find the vetical dop duing the time h y 1 gt m/s of flight: ( ) 1.6 m 3. (b) If the pitch speed is inceased the time the ball tavels is less, theefoe the dop distance deceases.

2 4. (c) Since the moon s gavity is less the dop distance deceases. Insight: The 1.6 m dop coesponds to about 5 ft, nealy the height of a playe. The 3 m/s (7 mi/h) speed is a bit slow; a sizzling fastball at 44 m/s (98 mi/h) would dop about 0.8 m, half as much. 1. Pictue the Poblem: The basketball s tajectoy is depicted at ight. Stategy: Use equations 4-7 to find the x and y positions of the basketball as a function of time. Use the ight tiangle fomed by the floo and the basketball s elease and landing points to wite a atio that allows us to calculate the time of flight and theefoe the initial height. Solution: 1. Find the y position as a function of time: y h 1 gt 0 h 1 gt. Find the x position as a function of time: x t 3. Use the tangent function fo the ight tiangle: 4. Now solve fo the flight time t: tanθ h 1 x gt gt t 5. Find the initial height: h 1 gt 1 t tanθ ( 4.0 m/s) tan s g 9.81 m/s 9.81 m/s ( ) 1.0 m ( ) s Insight: If the basketball playe thows the ball fom the same height but with a highe initial speed, the 30.0 angle will decease. Fo instance, 8.40 m/s poduces an angle of Dopping the ball fom est makes the angle Pictue the Poblem: The ball tavels along a paabolic ac, maintaining its hoizontal velocity but changing its vetical speed due to the constant downwads acceleation of gavity. Stategy: The given angle of the thow allows us to calculate the hoizontal component of the initial velocity by using the cosine function. The vetical component of the velocity can be found by using the sine function. The time it takes the acceleation of gavity to slow down the vetical speed, bing it to zeo, and speed it up again to its initial value is the same as the time the ball is in the ai. Solution: 1. (a) Find the x component of the initial velocity: x cosθ ( 17.0 m/s)cos m/s. (b) Find the y component of the initial velocity: y sinθ ( 17.0 m/s)sin m/s 3. Let v y y and use equations 4-6 to find the time of flight: t v y y g 9.75 ( 9.75 m/s) 1.99 s 9.81 m/s Insight: The flight of the ball is pefectly symmetic the angle of the motion is 35.0 below hoizontal at the instant it is caught, and the ball spends the same amount of time going upwad to the

3 peak of its flight as it does coming downwad fom the peak. This is only tue if the ball is caught at the same level fom which it was thown. 37. Pictue the Poblem: The tajectoy of the gil is depicted at ight. Stategy: Use equations 4-10 and the given time of flight, initial speed, and launch angle to detemine the initial height of the gil at the elease point. Solution: Use equations 4-10 to find the initial height of the gil at the elease point. If we let the elease height coespond to y 0, then the landing height is: y ( sinθ )t 1 gt ( )( sin 35.0 )( s) 1.5 m/s y 1.07 m ( 9.81 m/s ) s ( ) In othe wods, she was 1.07 m above the wate when she let go of the ope. Insight: The gil s speed upon impact with the wate is 5.10 m/s (check fo youself) o 11 mi/h. A fun plunge! 43. Pictue the Poblem: The baseball tavels along a paabolic ac and is caught at the same level fom which it was thown. Stategy: Because the ball lands at the same level fom which it was stuck, we can use the fomulas in section 4-5 to solve this poblem. We can use equation 4-1 to find the ball s initial speed and equation 4-11 to find the time of flight. Solution: 1. (a) Solve equation 4-1 fo :. (b) Solve equation 4-11 fo t: t g g R sin θ ( 9.81 m/s )( 96 ft m/ft) 9.8 m/s sin 90.0 ( 9.8 m/s) sinθ sin s 9.81 m/s Insight: Veify fo youself that the thow speed is 67 mi/h and the maximum height of the ball is.6 m.

4 45. Pictue the Poblem: The lava that eaches maximum altitude is huled staight upwad, comes to est momentaily, and falls staight downwad again. Stategy: Use equations 4-6 to detemine the launch speed equied to achieve the maximum altitude. Solution: 1. (a) Use equation 4-6 to find : v y y gδy 0 at the peak of flight y gδy 1.80 m/s ( )( m) 849 m/s. (b) If the launch speed emains the same, the maximum height of the ejected lava on Eath would be less than it is on Io because the acceleation of gavity on Eath is much geate than 1.80 m/s. Insight: On Eath a launch speed of 849 m/s would be.5 times the speed of sound oughly equal to the muzzle velocity of a ifle bullet! Tidal foces fom Jupite poduce these violent volcanic euptions on Io. 53. Pictue the Poblem: The tajectoy of the cok and its initial velocity as seen by an obseve on the gound ae depicted at ight. Stategy: Add the velocity of the balloon to the velocity of the cok as seen by an obseve in the balloon to find the velocity of the cok as seen by an obseve on the gound. Use the components of the initial velocity to find its magnitude and diection. Then use equations 4-10 to detemine the maximum height of the cok above gound and the time of flight. Solution: 1. (a) Add v bg (the velocity of the balloon elative to gound ) to v cb (the velocity of the cok elative to the balloon): v cb + v bg ( ) ˆx 5.00 m/s 5.00 m/s + (.00 m/s) ŷ ( ) ˆx + (.00 m/s) ŷ. (b) Use the components of v cg to find the speed: ( 5.00 m/s) + (.00 m/s ) 5.39 m/s 3. Use the components of v to find the diection: θ tan 1 y v x tan 1.00 m/s 5.00 m/s 1.8 above hoizontal

5 4. (c) Use equations 4-6 to find the maximum height of the cok, noting its initial height is 6.00 m: v y 0 v 0 y gδy Δy v 0 y g (.00 m/s) ( ) 0.04 m 9.81 m/s y max y 0 + Δy m 6.0 m 5. (d) Use equations 4-6 to find the time to each the peak of flight: t up v y y g 0.00 m/s 9.81 m/s 0.04 s 6. Use equations 4-6 to find the time to fall fom the peak of flight: y y 0 + y t 1 gt 0 y gt down ( ) t down y 0 g 6.0 m 9.81 m/s 1.14 s 7. Add the times to find the total time of flight: t flight t up + t down s 1.33 s Insight: Anothe way to solve pat (d) is to use v y y gδy to find v y, then use v y, y and the acceleation of gavity to find the total time of flight. It saves a step o two but doesn t eveal t up o t down.