Bob has just finished climbing a sheer cliff above a beach, and wants to figure

Question

Bob has just finished climbing a sheer cliff above a beach, and wants to figure out how high he climbed. All he has to use, however, is a baseball, a stopwatch, and a friend on the beach below with a long measuring tape. Bob is a pitcher and he knows that the fastest he can throw the bal is about vo 32.1 m/s. Bob starts the stopwatch as he throws the ball (with no way to measure the ball's initial trajectory), and watches carefully. The ball rises and then falls, and after ti - 0.910 s the ball is once again level with Bob. Bob cannot see well enough to time when the ball hits the ground. Bob's friend then measures that the ball landedx - 125 m from the base of the clif. How high up is Bob, if he hall started exacdly 2 m above the edge of the cliff? Bob's position TOOLS x10

Explanation / Answer

Time of flight, T = 2u sin(theta) /g

0.91 = 2 x 32.1 x sin(theta) /9.8

Theta = 7.98 degree

Tf = R (u cos (theta)) = 3.93 sec

Height, h = - (u sin(theta) t) + 0.5gt^2

= - (32.1 x sin7.98 x 3.93) + (0.5 x 9.8 x 3.93^2)

h = 58.17 m

Ball started 2 m above the edge,

So, H = 58.17 - 2 = 56.17 m

Comment in case any doubt please rate my answer....

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