Bob has just finished climbing a sheer cliff above a level beach and wants to figure out how high he climbed. All he has touse is a baseball, a stopwatch, and a friend on the ground below with a long measuring tape. Bob is a pitcher and knowsthat the fastest he can throw the ball is about vo = 32.1 m/s. Bob starts the stopwatch as he throws the ball, with no wayto measure the ball's initial trajectory, and watches carefully. The ball rises and then falls, and after t1 = 0.510 s the ball isonce again level with Bob. Bob cannot see well enough to time when the ball hits the ground. Bob's friend then measuresthat the ball hit the ground x =128 m from the base of the cliff. How high above the beach was the ball when itwas thrown?V.initial height: 27.1Incorrect

since given, height =v0=32.1 m/st1= 0.510 slet us suppose the angle subtended is θtherefore…

Answered: Bob has just finished climbing a sheer…

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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 ground with a long measuring tape. Bob is a pitcher, and knows that the fastest he can throw the ball is vo= 81.0 mph. 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 t₁ = 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 landed a distance of x = 442 ft from the base of the cliff. How high up is Bob if the ball started from exactly 5 ft above the edge of the cliff? 5 ft 1₁ height of the cliff: ft
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