Question

a ball is projected upward with an initial speed of approximately 50 m/s. the diagram at the right represents its position at 1-second intervals of time. at what location will the ball be moving downward with a speed of approximately 20 m/s? tap to toggle through answer options: d check answer

Explanation:

Step1: Recall projectile motion physics

The ball is projected upward with initial speed \( v_0 = 50 \, \text{m/s} \). The vertical speed as a function of time is \( v = v_0 - gt \), where \( g \approx 10 \, \text{m/s}^2 \). We need to find \( t \) when \( v = -20 \, \text{m/s} \) (downward, so negative).

Step2: Solve for time \( t \)

Using \( v = v_0 - gt \), substitute \( v = -20 \), \( v_0 = 50 \), \( g = 10 \):
\( -20 = 50 - 10t \)
\( 10t = 50 + 20 \)
\( 10t = 70 \)
\( t = 7 \) seconds.

Step3: Analyze position at \( t = 7 \)

The diagram has 1-second intervals. At \( t = 7 \), we check the positions. The vertical spacing (since horizontal motion is constant, vertical speed determines vertical position change). At \( t = 7 \), the ball’s vertical speed is downward 20 m/s, so its vertical position change over time should correspond to the location. Looking at the options, the location at \( t = 7 \) (7 seconds) would be where the vertical spacing and time interval match. From the diagram, the correct location (after 7 seconds, 1-second intervals) is the one corresponding to \( t = 7 \), which is the position labeled (or the one matching 7 intervals). Assuming the diagram’s labels, the correct location is the one at 7 seconds, which is the option (e.g., if D is at a certain time, but rechecking: wait, initial speed 50 m/s upward, time to reach max height: \( t_{max} = \frac{v_0}{g} = 5 \) seconds (at \( t = 5 \), speed 0). Then, after \( t = 5 \), it falls. At \( t = 7 \), time since max height is \( 7 - 5 = 2 \) seconds. Speed downward: \( v = gt = 10 \times 2 = 20 \, \text{m/s} \), which matches. So at \( t = 7 \), the position is 7 seconds after launch. The diagram’s points: each interval is 1 second. So the 7th interval (t=7) corresponds to the location. From the diagram, the correct option is the one at t=7, which is the location (e.g., if the options are labeled, and D is a candidate, but re-express: the ball’s speed downward 20 m/s at t=7 (since 5s up, 2s down, speed 20 m/s). So the position at t=7 is the 7th second, so the location is the one at 7 seconds, which is the correct answer (e.g., the option corresponding to t=7, likely the one with the correct vertical spacing for 7 seconds).

Answer:

The correct location is the one corresponding to 7 seconds after launch, which (from the diagram’s options) is the position at \( t = 7 \), likely the labeled option (e.g., if the diagram’s D is at t=7, but based on calculation, the answer is the location at 7 seconds, so the correct option is the one matching \( t = 7 \), which is the position with 7 intervals, so the answer is the location corresponding to \( t = 7 \) (e.g., the option labeled or positioned for 7 seconds, such as the one with vertical spacing matching 7 seconds of motion).