Bob has just finished climbing a sheer cliff above a beach, and wants to figure
ID: 1655673 • Letter: B
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 86.0 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 0.710 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 4.20 × 102 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?
Explanation / Answer
Time at/above launch height t = 2·Vo·sin/g
First, let's convert mph to fps:
86mph * 5280ft/mi * 1hr/3600s = 126.133 ft/s
0.710 s = 2 * 126.133ft/s * sin / 32.2ft/s²
0.710 = 7.83435 sin
= arcsin(0.710/7.83435) = 5.2º
Then the horizontal velocity is Vx = 126.133ft/s * cos5.2º = 125.614 ft/s
Then the ball was between launch height and the base of the cliff for time
t = x / v = 420ft / 125.614ft/s = 3.34 s
The initial vertical velocity is Vy = 126.133ft/s * sin5.2º = 11.432 ft/s
Upon returning to launch height, then, Vv = -11.432 ft/s
Then the vertical displacement from launch height to the base of the cliff is
y = Vv*t + ½at² = -11.432 ft/s * 3.34s - ½ * 32.2ft/s² * (3.34s)² = -217 ft
Therefore the cliff height is 217ft - 5ft = 212 ft.