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 ground below with a long measuring tape. Bob is a pitcher, and knows that the fastest he can throw the ball is about 31.7 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 0.510 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 121 m from the base of the cliff. How high up is Bob, if the ball started from exactly 2 m above the edge of the cliff?

Please providde steps and answer.

Answer 1

Use the eye-level arc to find the angle...
Equation of motion in the vertical direction
hf = hi + vi*t - 1/2*g*t^2

(hf - hi) = t*(vi - 1/2*g*t)

For this case, since it is level (hf - hi) = 0
0 = t*(vi - 1/2*g*t)

Solve for vi
0 = vi - 1/2*g*t
vi = 1/2*g*t

vi = 1/2*9.81 m/s^2 * 0.710 s = 3.48 m/s <--- This is vertical velocity

Definition of the vertical component of velocity:
vi = v*sin(theta)
theta = asin(vi/v)
theta = asin(3.48 / 31.7) = 6.3 deg

Definition of the horizontal component of velocity:
vh = v*cos(theta)
vh = 31.7 * cos(6.3) m/s = 31.5 m/s

Now we know how far the ball is from the base of the cliff. Use that to find the time it took to travel that distance.
t = distance / vh
t = 128 m / 31.5 m/s
t = 4.06 s

Back to the equation of motion in the vertical direction
hf = hi + vi*t - 1/2*g*t^2

Solve for hi, assume hf = 0
hi = 1/2*g*t^2 - vi*t
hi = 1/2*9.81 m/s^2 * (4.06 s)^2 - 3.48 m/s*4.06 s
hi = 66.7 m

Finally, the ball started 2m above the cliff.. so
cliff height = hi - 2
cliff height = 66.7 m - 2 m = 64.7 m