Catch/Overtake Problems Video Lessons

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# Problem: A rock is thrown vertically upward with a speed of 16.0 m/s. Exactly 1.00 s later, a ball is thrown up vertically along the same path with a speed of 23.0 m/s.(a) At what time will they strike each other?(b) At what height will the collision occur?Assuming that the order is reversed: the ball is thrown 1.00 s before the rock:(c) At what time will they strike each other?(d) At what height will the collision occur?

###### FREE Expert Solution

The problem requires us to determine the time and the height the two objects strike each other (meet), given their initial velocities

This is a motion of multiple objects problem involving vertical meet and catch. Whenever we're given this kind problem, we use the following steps:

1. Write the position equation for each object using UAM equation (3).
2. Set the position equations equal to each other.
3. Solve for time.
4. (If needed) Plug the time back into another equation to solve for Δy.

Recall that the four UAM equations are:

The displacement equation in the third UAM equation would be rewritten as a position equation because the two objects start at different times: it’s rewritten as:

$\overline{){{\mathbf{y}}}_{{\mathbf{f}}}{\mathbf{=}}{{\mathbf{y}}}_{{\mathbf{0}}}{\mathbf{+}}{{\mathbf{v}}}_{{\mathbf{0}}}{\mathbf{t}}{\mathbf{+}}\frac{\mathbf{1}}{\mathbf{2}}{{\mathbf{at}}}^{{\mathbf{2}}}}$

In this problem, we’re directly given some information:
· The initial speed of the rock, v0r = 16.0 m/s
· The initial speed of the ball, v0b = 23.0 m/s
As usual, we assume g = 9.8 m/s2.
We’ll set our coordinate system so that positive is up and the origin (y = 0) is at the base of the cliff.

The two are launched from the origin of the coordinate system  (same point). Thus, y0 for both objects is zero. When an object is launched from the origin, the final position is equal to the height/displacement from the launch point.

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###### Problem Details

A rock is thrown vertically upward with a speed of 16.0 m/s. Exactly 1.00 s later, a ball is thrown up vertically along the same path with a speed of 23.0 m/s.
(a) At what time will they strike each other?
(b) At what height will the collision occur?

Assuming that the order is reversed: the ball is thrown 1.00 s before the rock:
(c) At what time will they strike each other?
(d) At what height will the collision occur?