Holt Physics Problem 3E

Transcription

1 NAME DATE CLASS Holt Phyic Problem 3E PROJECTILES LAUNCHED AT AN ANGLE PROBLEM SOLUTION 1. DEFINE. PLAN A flying fih leap out of the water with a peed of Normally thee fih ue winglike fin to glide about 40 m before reentering the ocean, but in thi cae the fih fail to ue it wing and o only travel horizontally about 17.5 m. At what angle with repect to the water urface doe the fih leave the water? Ue the trigonometric identity (inq)(coq) = in (q) to olve for q. Given: Unknown:? Diagram: v i = 15.3 = 17.5 m g = 9.81 v i θ = 17.5 m Chooe the equation() or ituation: The horizontal component of the fih initial velocity, v x, i equal to the horizontal diplacement divided by the time of the jump. v x = v i (co q) = t The vertical diplacement of the fih i given by the equation for falling bodie, with the vertical component of the initial velocity, v y, ued. y = v y t 1 g t Becaue the fih land at the ame vertical poition from which it tarted, y = 0. y = 0 v y = v i (in q) = 1 g t Rearrange the equation() to iolate the unknown: Subtitute for t uing the equation for horizontal velocity. t = vi (co q) v i (in q) = 1 g vi (co q) (in q)(co q) = g vi Uing the trigonometric identity allow a olution for q to be found. (in q)(co q) = 1 [in (q)] in (q) = g vi Problem 3E Ch. 3 13

2 NAME DATE CLASS 3. CALCULATE in 1 g vi in 1 (9.81 )(17.5 m) (15.3 ) = 3.6 above the horizontal 4. EALUATE Subtituting the value for q into the original equation and olving for t produce a time of 1.5 for both, thu confirming the reult for q. ADDITIONAL PRACTICE 1. A baeball i thrown with an initial peed of If the ball horizontal diplacement i 17.6 m, at what angle with repect to the ground i the ball pitched? Ue the trigonometric identity (in q)(co q) = in (q) to olve for q.. A football i kicked o that it initial peed i 3.1. If the football reache a maximum height of 16.9 m, at what angle with repect to the ground i the ball kicked? 3. Jackie Joyner-Keree record long jump i 7.49 m. Suppoe he ran 9.50 to jump thi horizontal ditance. At what angle above the horizontal did he jump? Ue the trigonometric identity (in q)(co q) = in (q) to olve for q. 4. The mall jumping pider make up for their ize by their ability to leap relatively large ditance. Some can jump fifty time the length of their bodie. Suppoe a jumping pider leap a horizontal ditance of 18.5 cm with an initial peed of about 141 c. At what angle above the horizontal would a pider with thi peed have to leap in order to travel a range of 18.5 cm? Ue the trigonometric identity (in q)(co q) = in (q) to olve for q. 5. Olympic platform diver jump from a diving board that i 10.0 m above the water. Suppoe a diver jump from the board with an initial peed of The diver reache a maximum height of 11.7 m above the water, and land in the water at a horizontal ditance of 3.6 m from the end of the board. At what angle with repect to the board doe the diver leave the board? 6. A ball i thrown from a roof with a peed of 10.0 and an angle of 37.0 with repect to the horizontal. What are the vertical and horizontal component of the ball diplacement.5 after it i thrown? 7. A downed pilot fire a flare from a flare gun. The flare ha an initial peed of 50 and i fired at an angle of 35 to the ground. How long doe it take for the flare to reach it maximum altitude? Ch Holt Phyic Problem Bank

3 NAME DATE CLASS 8. In the port of ki jumping, a kier travel down the lope of a hill until he or he reache the takeoff. The takeoff i lanted lightly below the horizontal, o that the kier i able to travel in the air jut above the ground. Suppoe a kier leave the takeoff and land 73.0 m horizontally beyond the takeoff and 5.8 m below the takeoff. If the takeoff angle i 8.00 below the horizontal, what i the kier initial peed? 9. A hingle lide down a roof having a 30.0 pitch and fall off with a peed of.0. How long will it take to hit the ground 45 m below? 10. A hole at a miniature golf coure require the ball to roll up a ramp, fly over a mall tream, and then land on the green beyond the tream. The tream i 0.46 m wide, and the cup i 4.00 m beyond the tream edge. The ramp make an angle of 41.0 with the horizontal, and it upper edge i 0.35 m above the green. What mut the ball initial peed be in order for the ball to fly over the water and land directly in the cup? Problem 3E Ch. 3 15

4 Given Solution 9. v x = 1.50 v y,f = g y + v y,i y = m g = 9.81 v y,i = 0, o v y,f = v y = g y = v y = 70.0 v = v x + v y = = m / v = tan 1 v x vy 70.0 ()(9. 81 (. ) m) (1.50) + (70.0 ) = = tan from the vertical. 5 m / m / g y = v x x 10. v x = 85.3 t = y = 1.50 m = v x g = 9.81 g y = (85.3 ) ( range of arrow = 47. m ) ( m 0 m / ) = 47. m Additional Practice 3E 1. v i = 15.0 t = vi (co q) = 17.6 m y = v i (in q) t 1 g t = 0 v i (in q) = 1 g t = 1 g vi (co q) (in q)(co q) = g vi Uing the identity (in q)(co q) = in ( q), in ( q) = g vi in = (9.81 )( 17. 6m) in 1 g vi x (15. 0 ). v i = 3.1 v y,f v y,i = g y y max = 16.9 m At maximum height, v y,f g = v y,i = v i (in q) = g y max g ym ax vi in 1 = ()(9. 81 (16. ) 9m) in Section Five Problem Bank Ch. 3 11

5 Given Solution 3. = 7.49 m Uing the form of the equation derived in problem 1, v i = 9.50 (in q)(co q) = in( q) = g x v i g = 9.81 g vi x in 1 = in 1 (9.81 )(7.49m) ( v i = 141 c Uing the form of the equation derived in problem 1, = 18.5 cm (in q)(co q) = in( q) = g vi g = 9.81 in 1 in 1 (9.81 )( m) ( ) 33.0 g vi x 5. v i = 6.03 v y,f v y,i = g y h i = 10.0 m At maximum height, v y,f = 0. h max = h f = 11.7 m v y,i = v i (in q) = g y max = 3.6 m g = 9.81 For the diver, h f i the maximum height above the diving board. y = h f h i g (hf h i ) vi ()(9. in 1 81) = (11. ) 7m 10. 0m) in v i = 10.0 = v i (co q) t = (10.0 )(co 37.0 )(.5 ) 37.0 = m t =.5 y = v i (in q) t 1 g t = (10.0 )(in 37.0 )(.5 ) 1 (9.81 )(.5 ) g = 9.81 = 15 m 31 m y = 16 m 7. v i = 50 At the maximum height 35 v y,f = v y,i g t = 0 g = 9.81 v y,i = v i (in q) = g t t = v i(in q) g ) (in 35 ) 9.81 t = 15 Ch. 3 1 Holt Phyic Solution Manual

6 Given Solution 8. = 73.0 m t = vi (co q) y = 5.8 m y = v i (in q) t 1 g t = v i (in q) 8.00 vi (co q) g = 9.81 g y =(tanq) vi ( co q) g v i ( co q) =(tan q) y v i (co q) g = [(ta n q) y] 1 g vi (c o q) g v i = (co q) [(tan q) y] (9.81 v i = )(73.0 m) ()[co( 8.00 )] [(73.0 m)(tan[ 8.00 ]) ( 5.8 m)] (9.81 v i = )(73.0 m) ()[co( 8.00 )] ( 10.3 m m) (9.81 v i = )(73.0 m) ()[ co( 8.00 )] (4.5 m) v i = y = v i (in q) t 1 g t v i =.0 g t [v i (in q)] t + y = 0 y = 45 m g = 9.81 Solving for t uing the quadratic equation, v i (in q) ± t = [ v i ( in q)] 4 g ( y) g (.0 )[in( 30.0 )] ± [(. 0)[i n( 30.0 ) ] () (9.81 ( 45 ) m) t = ± 1. 0m / m / 1.0 ± m / t = = ± t = 9.81 t mut be poitive, o the poitive root mut be choen. 9 t = = 3.0 Section Five Problem Bank Ch. 3 13

7 Given Solution = 0.46 m tot = 1 + = 4.00 m t = vi (co q) 41.0 y = v i (in q) t 1 g t = v i (in q) xi + v i ( co q) y = 0.35 m y = ( 1 + )(tan q) g ( + ) vi 1(c o q) g = g v x1 + i( co q) g( v i = 1 + ) (co q) [( 1 + )(tan q) y] (9.81 v i = )(0.46 m m) ()(co 41.0 ) [(0.46 m m)(tan 41.0 ) ( 0.35 m)] (9.81 v i = )(4.46 m) ()(co 41.0 ) (3.88 m m) (9.81 v i = )(4.46 m) ()(co 41.0 ) (4.3 m) v i = 6.36 Additional Practice 3F 1. v bw = 58.0 kh, forward v be = v bw + v we =+58.0 kh + ( 55.0 kh) =+3.0 kh =+58.0 kh 1.4 km t = = v we = 55.0 kh, backward v be 3.0 kh = 55.0 kh = 1.4 km t = 0.47 h = 8 min. v we = 1.50, wet v me = v mw + v we = = 5.70, wet v mw = 4.0, wet time of travel with walkway: = m t 1 = = m = 149 v me 5.70 time of travel without walkway: t = = m = 0 v mw 4.0 time aved = t t 1 = = v 1e = 86 kh, forward v 1 + v e = v 1e v e = 5 kh, forward v 1 = v 1e v e = km v 1 = v 1e v e = 86 kh 5 kh = 34 kh t = km =. 10 h v 1 34 k h t = (. 10 h) h = 79 Ch Holt Phyic Solution Manual