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 ????0=34.1 m/s.v0=34.1 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 ????1=0.510 st1=0.510 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 ????=126 mx=126 m 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

Answer:

56.0 m

Explanation:

We know that after 0.510 s, the ball is level with Bob again.  We can use this to find the vertical component of the initial velocity.

y = y₀ + v₀ᵧ t + ½ gt²

h+2 = h+2 + v₀ᵧ (0.510) + ½ (-9.8) (0.510)²

v₀ᵧ = 2.50 m/s

Since the magnitude is 34.1 m/s, we can now find the horizontal component:

v₀² = v₀ₓ² + v₀ᵧ²

(34.1)² = v₀ₓ² + (2.50)²

v₀ₓ = 34.0 m/s

And since we know the ball lands 126 m from the base of the cliff, we can find the time it takes to land:

x = x₀ + v₀ₓ t + ½ at²

126 = 0 + (34.0) t + ½ (0) t²

t = 3.71 s

Finally, we can now find the height of the cliff:

y = y₀ + v₀ᵧ t + ½ gt²

0 = h+2 + (2.50) (3.71) + ½ (-9.8) (3.71)²

h = 56.0 m