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:

__Write the__for each object using UAM equation (3).**position equation**- Set the position equations
__equal to each other__. - Solve for
.**time** - (If needed) Plug the
**time**__back into another equation__to solve for Δy.

Recall that the four **UAM **equations are:

$\overline{){\mathbf{}}{\mathbf{\left(}}{\mathbf{1}}{\mathbf{\right)}}{\mathbf{}}{{\mathit{v}}}_{{\mathit{f}}}{\mathbf{=}}{{\mathit{v}}}_{{\mathbf{0}}}{\mathbf{-}}{\mathit{g}}{\mathit{t}}\phantom{\rule{0ex}{0ex}}\mathbf{}{\mathbf{\left(}}{\mathbf{2}}{\mathbf{\right)}}{\mathbf{}}{\mathbf{\u2206}}{\mathit{y}}{\mathbf{=}}\mathbf{\left(}\frac{{\mathit{v}}_{\mathit{f}}\mathbf{+}{\mathit{v}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{\right)}{\mathit{t}}\phantom{\rule{0ex}{0ex}}\mathbf{}{\mathbf{\left(}}{\mathbf{3}}{\mathbf{\right)}}{\mathbf{}}{\mathbf{\u2206}}{\mathit{y}}{\mathbf{=}}{{\mathit{v}}}_{{\mathbf{0}}}{\mathit{t}}{\mathbf{-}}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{g}}{{\mathit{t}}}^{{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}\mathbf{}{\mathbf{\left(}}{\mathbf{4}}{\mathbf{\right)}}{\mathbf{}}{{{\mathit{v}}}_{{\mathit{f}}}}^{{\mathbf{2}}}{\mathbf{=}}{{{\mathit{v}}}_{{\mathbf{0}}}}^{{\mathbf{2}}}{\mathbf{-}}{\mathbf{2}}{\mathit{g}}{\mathbf{\u2206}}{\mathit{y}}{\mathbf{}}}$

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, v_{0r} = 16.0 m/s

· The initial speed of the ball, v_{0b} = 23.0 m/s

As usual, we assume g = 9.8 m/s^{2}.

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, y_{0} 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.

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?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Catch/Overtake Problems concept. You can view video lessons to learn Catch/Overtake Problems. Or if you need more Catch/Overtake Problems practice, you can also practice Catch/Overtake Problems practice problems.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Ye's class at MSSTATE.