Question

10. A scientist throws a rock straight upwards on Mars and
it reaches a height of 55m.
a. What was the initial velocity of the rock?
| b. How long was the rock in the air for?

Answer 1

a. 20.4 m/s upward

The motion of the rock thrown upward is a free-fall motion, with a constant acceleration downward, given by

$g=-3.8 m/s^2$ (acceleration of gravity on Mars)

So, we can use the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity  of the rock

u is the initial velocity

a is the acceleration  ($g=-3.8 m/s^2$)

s is the displacement

For the rock in the problem:

v = 0 (at the maximum height, the velocity is zero)

s = 55 m (the maximum height)

Solving for u, we find the initial velocity:

$u=\sqrt{v^2-2as}=\sqrt{0-2(-3.8)(55)}=20.4 m/s$

b. 10.74 s

First of all, we can find the time the rock takes to reach the maximum height, that is given by the equation

$v=u+at$

And solving for t,

$t=\frac{v-u}{a}=\frac{0-20.4}{-3.8}=5.37 s$

Now we notice that the motion of the rock is symmetrical, so the way down takes exactly the same as the way up: therefore, the total time of flight of the rock is

$T=2t=2(5.37)=10.74 s$