A man on the edge of a cliff H = 50.0 m high throws a ball directly upward. It r

Question

A man on the edge of a cliff H = 50.0 m high throws a ball directly upward. It returns past him 1.7 s later. H actually is the height of the point of release of the ball above the base of the cliff. You may neglect air resistance and assume that the motion is described as one of constant acceleration of g = 9.81 m/s2 in the downward direction. How much time does it take for the ball to reach its maximum height? How high above the point of release did the ball go? What is the speed of the ball when it passes him on the way back down? How long after the man throws it does the ball hit the ground at the bottom of the cliff? What is the speed of the ball when it hits the ground? The man throws now throws another ball straight up and it returns past him 3.4 seconds later, i.e., the second ball takes twice as long as the first ball to return past its release point. How does h2, the maximum height above the point of release, and v2, the speed of the ball as it passes the point of release, compare to the original values, h1 and v1? h2 = 2h1 and v2 = 2v1 h2 = 2h1 and v2 > 2v1 h2 > 2h1 and v2 = 2v1 h2 > 2h1 and v2 > 2v1

Explanation / Answer

1. 0.85 s
2.
0 = ut -1/2 gt2 => u = 1/2 gt = 0.5*9.81*1.7 = 8.3385 m/s

s = 8.3358*0.85-0.5*9.81*0.85*0.85 = 3.544 m

3.

speed = - 8.3385 m/s

4.

-50 = 8.3385* t -0.5*9.81*t^2

=> t = 4.154 seconds

5.

v2 = u2 + 2as = 8.3385*8.3385 +9.81*2*50

=> v = 32.41 m/s

6.

0 = u't-1/2 gt^2 => u' = 1/2gt = 0.5*9.81*3.4 => u' = 2u

h' =2*8.3385*1.7 -0.5*9.81*1.7*1.7 = 14.17 m

=> h' > 2 h

therefore option c is correct

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