2_D_Motion_Problems_Solutions

# 2_D_Motion_Problems_Solutions - Physics 110 Spring 2006 2-D...

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Physics 110 Spring 2006 2-D Motion Problems: Projectile Motion – Their Solutions 1. A place-kicker must kick a football from a point 36 m (about 40 yards) from the goal, and half the crowd hopes the ball will clear the crossbar, which is 3.1m high. When kicked the all leaves the ground with a speed of 20 m/s at an angle of 53 0 to the horizontal. a. Does the ball clear or fall short of the crossbar? b. Does the ball approach the crossbar while still rising or while falling? (a) From our equations of motion, the horizontal velocity is constant. This gives us the flight time for any horizontal distance starting with initial x velocity v i cos θ . Thus the vertical height of the trajectory is given as y = x tan θ i i i v gx 2 2 cos 2 2 . With x = 36.0 m, v i = 20.0 m/s, and = 53.0°, we find y = (36.0 m)(tan 53.0°) – (9.80 m/s 2 )(36.0 m) 2 (2)(20.0 m/s) 2 cos 2 53.0° = 3.94 m. The ball clears the bar by (3.94 – 3.05) m = 0.889 m . (b) The time the ball takes to reach the maximum height is t 1 = v i sin q i g = (20.0 m/s)(sin 53.0°) 9.80 m/s 2 = 1.63 s. The time to travel 36.0 m horizontally is t 2 = x v ix , which gives t 2 = 36.0 m (20.0 m/s)(cos 53.0°) = 2.99 s. Since t 2 > t 1 the ball clears the goal on its way down . 2. A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8m/s at an angle of 20 0 below the horizontal, where it strikes the ground 3 seconds later. a. How far horizontally from the base of the building does the ball strike the ground? b. At what height was the ball thrown? c. How long does it take the ball to reach a point 10m below the level of launching? (a) x = v xi t = (8.00 cos 20.0°)(3.00) = 22.6 m (b) Taking y positive downwards, y = v yi t + 1 2 gt 2 = 8.00(cos 20.0°)3.00 + 1 2 (9.80)(3.00) 2 = 52.3 m

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(c) 10.0 = 8.00 cos 20.0° t + 1 2 (9.80) t 2 , which gives a quadratic in t. The solutions to 4.90 t 2 + 2.74 t – 10.0 = 0 are given by the quadratic formula, t = –2.74 ± (2.74) 2 + 196 9.80 = 1.18 s 3. A firefighter 50m away from a burning building directs a stream of water from a fire hose at an angle of 30 0 above the horizontal as shown below. If the speed of the stream is 40m/s, at what height will the water strike the building? From our equations of motion, the horizontal velocity is constant. This gives us the flight time for any horizontal distance starting with initial x velocity v i cos θ . Thus the vertical height of the trajectory is given as y = x tan θ i i i v gx 2 2 cos 2 2 . Substituting the known quantities, we find h = 18.7m. 4. As some molten metal splashes off of the ground, one droplet flies off to the east with an initial speed v i at an angle i above the horizontal while the other drop flies off to the west with the same speed and at the same angle above the horizontal. In terms of v i and i , find the distance between the droplets as a function of time. At any time t , the two drops have identical y -coordinates. The distance between the two drops is then just twice the magnitude of the horizontal displacement either drop has undergone. Therefore, d = 2| x ( t )| = 2( v xi t ) = 2( v i cos i ) t = i i t v cos 2 .
5. A projectile is fired up an incline (incline angle φ ) with an initial speed v i

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2_D_Motion_Problems_Solutions - Physics 110 Spring 2006 2-D...

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