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 32.5 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 then and then falls, and after .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 125 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?
The correct answer and explanation is:
Problem Analysis
Bob’s goal is to determine the height of the cliff he climbed using physics principles. The problem involves analyzing the motion of the baseball in both horizontal and vertical directions. This requires splitting the motion into components and using kinematic equations.
Step-by-Step Solution
- Horizontal Motion:
- The ball’s horizontal velocity (vxv_x) remains constant.
- Using the formula vx=distancetimev_x = \frac{\text{distance}}{\text{time}}: vx=125 mttotalv_x = \frac{125 \, \text{m}}{t_{\text{total}}} Here, ttotalt_{\text{total}} is the total time the ball is in the air.
- Vertical Motion:
- The vertical velocity (vyv_y) changes due to gravity.
- When the ball is level with Bob after t=0.510 st = 0.510 \, \text{s}, it has reached its highest point and started descending.
- The vertical component of the initial velocity (vy0v_{y0}) is determined by: vy0=g⋅tv_{y0} = g \cdot t where g=9.8 m/s2g = 9.8 \, \text{m/s}^2. Substituting t=0.510 st = 0.510 \, \text{s}: vy0=9.8⋅0.510=5.0 m/sv_{y0} = 9.8 \cdot 0.510 = 5.0 \, \text{m/s}
- Initial Velocity:
- The magnitude of the initial velocity is v0=32.5 m/sv_0 = 32.5 \, \text{m/s}. Using Pythagoras’ theorem: vx2+vy02=v02v_x^2 + v_{y0}^2 = v_0^2 Solving for vxv_x: vx=32.52−52=32.1 m/sv_x = \sqrt{32.5^2 – 5^2} = 32.1 \, \text{m/s}
- Total Time:
- Horizontal velocity is vx=32.1 m/sv_x = 32.1 \, \text{m/s}. Using ttotal=dvxt_{\text{total}} = \frac{d}{v_x}: ttotal=12532.1≈3.89 st_{\text{total}} = \frac{125}{32.1} \approx 3.89 \, \text{s}
- Vertical Displacement:
- The vertical motion equation is: y=vy0t+12gt2y = v_{y0} t + \frac{1}{2} g t^2 Substituting t=3.89 st = 3.89 \, \text{s} and vy0=5 m/sv_{y0} = 5 \, \text{m/s}: y=5(3.89)−12(9.8)(3.892)y = 5(3.89) – \frac{1}{2}(9.8)(3.89^2) y=19.45−73.88≈−54.43 my = 19.45 – 73.88 \approx -54.43 \, \text{m}
- Cliff Height:
- Since the ball was thrown from 2 m above the cliff edge, the cliff height is: Cliff Height=∣y∣+2=54.43+2=56.43 m\text{Cliff Height} = |y| + 2 = 54.43 + 2 = 56.43 \, \text{m}
Final Answer
Bob climbed approximately 56.4 m.
Explanation
The problem combines horizontal motion (constant velocity) and vertical motion (accelerated due to gravity). By knowing the ball’s speed, horizontal distance, and flight time, we calculate the vertical height from which the ball was thrown. Using kinematic equations and breaking the motion into components ensures accurate results.