Question

a ball is projected upward with an initial speed of approximately 40 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 10 m/s? tap to toggle through answer options: tap here

Explanation:

To solve this, we analyze the motion of the ball projected upward:

Step 1: Recall Free - Fall Motion Concepts

A ball projected upward undergoes free - fall (assuming negligible air resistance). The acceleration due to gravity \(g\approx10\ m/s^{2}\) acts downward. The velocity of the ball at any time \(t\) is given by \(v = v_0 - gt\), where \(v_0 = 40\ m/s\) (initial upward velocity) and \(v\) is the velocity at time \(t\). When the ball is moving downward, \(v\) is negative (if we take upward as positive). We want to find the time when \(v=- 10\ m/s\) (downward speed of \(10\ m/s\)).

Using the formula \(v=v_0 - gt\), substitute \(v = - 10\ m/s\), \(v_0 = 40\ m/s\) and \(g = 10\ m/s^{2}\):

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Step 2: Analyze the Position - Time Diagram

The ball is at 1 - second intervals. At \(t = 0\) (launch), let's assume the initial position. As time increases, the ball moves up, slows down, reaches maximum height (where velocity is 0), then moves down, speeding up.

We know that the time to reach maximum height: at maximum height, \(v = 0\). Using \(v=v_0 - gt\), \(0 = 40-10t\), so \(t = 4\ s\) (time to reach max height). Then, after \(t = 4\ s\), the ball starts moving down. At \(t=5\ s\) (1 second after reaching max height), the ball is moving down.

Looking at the diagram, we analyze the positions. The key is to match the time \(t = 5\ s\) with the position. From the diagram's structure (with points at 1 - second intervals), the position corresponding to \(t = 5\ s\) (when the ball has a downward speed of \(10\ m/s\)) is at the location corresponding to that time interval. By analyzing the vertical motion and the grid (where each interval is 1 second), the ball's position at \(t = 5\ s\) (moving downward with \(10\ m/s\)) is at the point that is 5 seconds after launch. From the diagram, this position is at the location of point \(G\) (or the relevant point in the diagram that corresponds to \(t = 5\ s\) of motion, considering the 1 - second intervals and the trajectory).

Answer:

The ball will be at the location corresponding to \(t = 5\ s\), which is the position of point \(G\) (or the appropriate labeled point in the diagram that matches the 5 - second mark of motion with downward speed of \(10\ m/s\)). If we assume the diagram's points are labeled such that the 5 - second position is \(G\), the answer is the position of \(G\).