Question

question 4 of 15
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 ( v_0 = 86.0 ) mph. bob starts the stopwatch as he throws the ball upward and outward (with no way to measure the ball’s initial trajectory) and watches carefully.

after ( t_1 = 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 measures that the ball landed a distance of ( x = 376 ) ft from the base of the cliff.

the gravitational acceleration in imperial units is ( 32.2 ) ft/s².

how tall ( h_c ) is the cliff if bob threw the ball from exactly 5 ft above the cliff edge?

( h_c = ) (\boxed{}) ft

Explanation:

Step1: Convert initial velocity to ft/s

First, convert \( v_0 = 86.0 \) mph to ft/s. We know that 1 mile = 5280 ft and 1 hour = 3600 s. So, \( v_0 = 86.0 \times \frac{5280}{3600} \) ft/s. Calculating that: \( 86.0 \times \frac{5280}{3600} \approx 126.133 \) ft/s.

Step2: Find vertical component of initial velocity

The time to reach the same height (level with Bob) is \( t_1 = 0.910 \) s. For projectile motion, the time to go up and come back to the same height is \( t_1 = \frac{2v_{0y}}{g} \), where \( v_{0y} \) is the vertical component of initial velocity and \( g = 32.2 \) ft/s². Solving for \( v_{0y} \): \( v_{0y} = \frac{g t_1}{2} \). Plugging in values: \( v_{0y} = \frac{32.2 \times 0.910}{2} \approx 14.651 \) ft/s.

Step3: Find horizontal component of initial velocity

Using Pythagoras, \( v_{0x} = \sqrt{v_0^2 - v_{0y}^2} \). Plugging in \( v_0 \approx 126.133 \) ft/s and \( v_{0y} \approx 14.651 \) ft/s: \( v_{0x} = \sqrt{(126.133)^2 - (14.651)^2} \approx \sqrt{15909.5 - 214.66} \approx \sqrt{15694.84} \approx 125.28 \) ft/s.

Step4: Find total time of flight

The horizontal distance \( x = 376 \) ft. In horizontal motion (constant velocity), \( x = v_{0x} t \), so total time \( t = \frac{x}{v_{0x}} \). Plugging in: \( t = \frac{376}{125.28} \approx 3.001 \) s.

Step5: Analyze vertical motion to find cliff height

Bob throws from 5 ft above cliff edge. Let's take upward as positive. The vertical displacement from Bob's position (5 ft above cliff) to the ground is \( - (h_c + 5) \) ft (since ground is below). The vertical motion equation is \( y = v_{0y} t - \frac{1}{2} g t^2 \). Here, \( y = - (h_c + 5) \), \( t \approx 3.001 \) s, \( v_{0y} \approx 14.651 \) ft/s, \( g = 32.2 \) ft/s².

Substitute into equation: \( - (h_c + 5) = 14.651 \times 3.001 - \frac{1}{2} \times 32.2 \times (3.001)^2 \)

Calculate right side: \( 14.651 \times 3.001 \approx 43.968 \); \( \frac{1}{2} \times 32.2 \times 9.006 \approx 16.1 \times 9.006 \approx 144.997 \)

So, \( - (h_c + 5) = 43.968 - 144.997 \approx -101.029 \)

Multiply both sides by -1: \( h_c + 5 = 101.029 \)

Subtract 5: \( h_c = 101.029 - 5 \approx 96.0 \) ft (rounded appropriately)

Answer:

\( \boxed{96.0} \) (or similar value depending on rounding during steps)