Question

The acceleration of gravity near the surface of Mars is 3.72m/sec^2. A rock is thrown straight up from the surface with an initial velocity of 23m/sec. How high does it go? (Hint: When is the velocity zero?) Thank you!!!

Answer 1

`color(blue)(71.02m)`

Explanation:

When the rock is projected upwards it will immediately start decelerating due to the opposing force of gravity. When the velocity reaches zero on the ascent the rock will be at its highest point, it will then fall back down.

Will will use the equation:

`v^2=u^2+2as`

Where:

`bbv` in final velocity, `bbu` is initial velocity, `bba` is acceleration and `bbs` is the distance.

We have:

`v=0color(white)(88)` highest point

`u=23ms^(-1)`

`a=-3.72ms^(-2)color(white)(88)` this is negative since we are decelerating

Plugging these in the equation:

`0=(23)^2+2(-3.72)s`

Rearranging to find `bbs`:

`s=(-(23)^2)/(2(-3.72))=529/7.44=52900/744=71.10m` 2 d.p.