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

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 ball is about 𝑣0=33.7 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 𝑡1=0.710 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 𝑥=127m from the base of the cliff. How high up is Bob, if the ball started exactly 2 m above the edge of the cliff?

Answer 1

To determine how high Bob climbed, we need to first determine how long it took for the ball to reach its maximum height. We know that the time it takes for the ball to reach its maximum height and the time it takes for the ball to fall back to its initial height are equal. Therefore, it took the ball 𝑡1/2 = 0.710/2 = 0.355 s to reach its maximum height.

Using the kinematic equation:

𝑥 = 𝑣0𝑡 + 1/2 𝑎𝑡^2

where 𝑣0 is the initial velocity, 𝑎 is the acceleration due to gravity, and 𝑡 is the time.

We can find the maximum height (ℎ) the ball reached above the cliff by setting 𝑡 = 𝑡1/2 and solving for ℎ:

ℎ = 𝑣0(𝑡1/2) - 1/2 𝑔(𝑡1/2)^2

= (33.7 m/s)(0.355 s) - 1/2 (9.81 m/s^2)(0.355 s)^2

= 5.34 m

So, the ball reached a maximum height of 5.34 meters above the edge of the cliff.

To determine the height of the cliff (ℎ𝑐𝑙𝑖𝑓𝑓), we can use the kinematic equation again:

𝑥 = 𝑣0𝑡 + 1/2 𝑎𝑡^2

where 𝑣0 is the initial velocity, 𝑎 is the acceleration due to gravity, and 𝑡 is the time it took for the ball to land on the beach.

Since we don't know the time it took for the ball to land, we need to use the fact that the total time of flight (𝑡𝑡𝑜𝑡𝑎𝑙) is twice the time it took for the ball to reach its maximum height:

𝑡𝑡𝑜𝑡𝑎𝑙 = 2(𝑡1/2) = 2(0.355 s) = 0.71 s

Using the total time of flight and the horizontal distance traveled by the ball, we can solve for the height of the cliff:

𝑥 = 𝑣0(𝑡𝑡𝑜𝑡𝑎𝑙) + 1/2 𝑎(𝑡𝑡𝑜𝑡𝑎𝑙)^2

𝑥 = 𝑣0(0.71 s) + 1/2 (9.81 m/s^2)(0.71 s)^2

ℎ𝑐𝑙𝑖𝑓𝑓 = 𝑥 + 2 m - ℎ

= 127 m + 2 m - 5.34 m

= 123.66 m

Therefore, Bob climbed a cliff with a height of approximately 123.66 meters.