Question

An astronaut on the surface of an unexplored planet gently throws a rock upwards. It takes $4.00$ sec for the rock to reach a maximum height of $24.00m$ relative to where it was released before coming back down.
If we take gravitational acceleration to be constant, what is the planet's gravity?

• A. $-3.00m/s^2$
• B. $-2.00m/s^2$
• C. $-1.00m/s^2$
• D. $-0.50m/s^2$
• E. Cannot be determined

Answer 1

Answer: (A)

$-3.00m/s^2$

Since it is given that acceleration is constant, we can apply our standard kinematic equations.

using the position function, we can get an equation that gives two unknowns: acceleration and the initial velocity

$y\left(t\right)=\frac{1}{2}a{t}^2+v_{0}t+y_0$

$24.00m=\frac{1}{2}a{\left(4.00s\right)}^2+v_0\left(4.00s\right)$

to solve this, you need torecognize that in addition, when the rock reaches its maximum height,its velocity is momentarily zero.

thus, $v\left(t\right)=0.00m/s=at+v_0=a\left(4.00s\right)+v_0$

This gives us two euations and two unknowns, which you can solve via substitution to get option A, $a=-3.00\frac{m}{s^2}$