uestion 9 of 12 Sapling Learning Bob has just finished cimbing a sheer ciff abov
ID: 1579201 • Letter: U
Question
uestion 9 of 12 Sapling Learning Bob has just finished cimbing a sheer ciff 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 ground measuring tape. Bob is a pitcher, and knows that the fastest he can throw the ball is about 34.9 m/s. Bob a long as he no to watches carefully. The ball rises and then falls, and after 0.910 seconds the ball is once again level with Bob. Bob can't see well enough to time when the ball hits the ground. Bob's friend then measures that the ball landed 129 m from the base of the cif. How high up is Bob, if the ball started from exactly 2 m above the edge of the cliff? Im 0910 s 2 m 129 mExplanation / Answer
In vertical,
yf - yi = v0y t + ay t^2 / 2
0 - 0 = (34.9 sin(theta))(0.910) - (9.8 x 0.910^2 / 2)
sin(theta) = 0.128
theta = 7.34 deg
in horizontal,
t = 129 / (34.9 cos7.34) = 3.73 sec
in vertical,
0 - (2 + h) = (34.9 sin(7.34))(3.73) - (9.8 x 3.73^2 / 2)
2 + h = 51.5
h = 49.5 m
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