Question

a robot probe drops a camera off the rim of a 239 m high cliff on mars, where the free - fall acceleration is - 3.7 m/s². find the velocity with which the camera hits the ground. find the time required for it to hit the ground.

Explanation:

Step1: Identify the relevant kinematic - equation

The kinematic equation for vertical motion under constant acceleration is $y = y_0+v_0t+\frac{1}{2}at^{2}$, where $y - y_0$ is the displacement, $v_0$ is the initial velocity, $t$ is the time, and $a$ is the acceleration. Here, $y - y_0=- 239$ m (taking downwards as negative), $v_0 = 0$ m/s, and $a=-3.7$ m/s².

Step2: Substitute values into the equation

Substituting into $y - y_0=v_0t+\frac{1}{2}at^{2}$, we get $-239=0\times t+\frac{1}{2}\times(- 3.7)t^{2}$.

Step3: Solve for time $t$

First, simplify the equation to $-239 = - 1.85t^{2}$. Then, $t^{2}=\frac{239}{1.85}\approx129.19$. Taking the square - root of both sides, $t=\sqrt{129.19}\approx11.37$ s.

Answer:

$t\approx11.4$ s