A fielder throws a cricket ball straight up at 18 m/s to celebrate a catch…

Contributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A fielder throws a cricket ball straight up at $18\ \text{m/s}$ to celebrate a catch. Ignoring the air:

How long does it take to reach the top?
How high does it rise above his hand?
When does it come back to his hand?

Need a reminder? The key ideaShow

Free fall is steady acceleration with $a = -g$ (up positive), $g = 9.8\ \text{m/s}^2$ :

\displaystyle v = u - gt

\displaystyle y = ut - \tfrac12 gt^2

\displaystyle v^2 = u^2 - 2gy

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

← All Problems

← A flowerpot falls from rest off a balcony 12 m above the pavement…From the top of a cliff 45 m above the sea, a boy throws a stone straight up at 12 m/s… →

More onMechanics › Motion in a Straight Line Mechanics › Gravitation

Discussion (0)

Loading…