Question

On Mars, where air resistance is negligible, an astronaut drops a rock from a cliff and notes that the rock falls about d meters during the first t seconds of its fall. Assuming the rock does not hit the ground first, how far will it fall during the first 4t seconds of its fall?

Answer 1

The distance covered by the rock during the $4t$ seconds of its fall will be $\fbox{16d}$.

Explanation:

Given:

Distance covered by the rock in $t\text{ sec}$ is $d$.

Concept:

As the astronaut drops the rock from the top of a cliff, the stone falls freely under the acceleration due to gravity of the Mars. This motion of the rock on the surface of the mass occurs according to the second equation of motion.

$\boxed{S=v_it+\dfrac{1}{2}g_mt^2}$

Here, $S$ is the distance covered by the rock, $v_i$ is the initial velocity of the rock, $g_m$ is the acceleration due to gravity on the surface of Mars and $t$ is the time for which the rock falls.

Since the rock is dropped from the top of the cliff, the rock will not have any initial velocity. So, the initial velocity of the rock will be zero.

Substitute the values in the given equation.

$d=0(t)+\dfrac{1}{2}g_mt^2\\d=\dfrac{1}{2}g_mt^2$

Now, in order to find the distance covered by the rock as it falls for time $4t$, substitute $4t$ for $t$ in the above expression.

$\begin{aligned}d'&=0(t)+\dfrac{1}{2}g_m(4t)^2\\&=16.\dfrac{1}{2}g_mt^2\end{aligned}$

Substitute $d$ for $\dfrac{1}{2}g_mt^2$ in above expression.

$d'=16.d$

Thus, The distance covered by the rock during the $4t$ seconds of its fall will be $\fbox{16d}$.

Learn More:

1. Effect on the acceleration while sliding down the hill brainly.com/question/2286502

2. Expression for the acceleratuon of the block under friction brainly.com/question/6088121

3. Magnitude of acceleration of the car brainly.com/question/6423792

Answer Details:

Grade: High School

Subject: Physics

Chapter: Acceleration

Keywords:

Mars, cliff, stone, acceleration, gravity, falls, rock, top, time, t, 4t, distance, ground, equation of motion, initial velocity.

Answer 2

Answer:

$d_1 = 16 d$

Explanation:

As we know that initial speed of the fall of the stone is ZERO

$v_i = 0$

also the acceleration due to gravity on Mars is g

so we have

$d = v_i t + \frac{1}{2}gt^2$

now we have

$d = 0 + \frac{1}{2}g t^2$

now if the same is dropped for 4t seconds of time

then again we will use above equation

$d_1 = 0 + \frac{1}{2}g(4t)^2$

$d_1 = 16(\frac{1}{2}gt^2)$

$d_1 = 16 d$