An astronaut on a distant planet wants to determine its acceleration due to gravity. The astronaut throws a rock straight up with a velocity of + 15 m/s and measures a time of 20.0 s before the rock returns to his hand. What is the acceleration (magnitude and direction) due to gravity on this planet?

An Astronaut On A Distant Planet Wants To Determine

This problem aims to find the acceleration due to the gravity of an object on a distant planet. The concepts required to solve this problem are related to gravitational physics, which include Newton’s equations of gravitational motion.

A motion under the influence of gravity directs to the vertical movement of an object whose motion is impacted by the existence of gravity. Whenever an object falls, a force attracts that object downwards known as gravity.

Newton’s equations of motion are related to an object moving in a horizontal direction, which means there is no gravitational acceleration imposed on the object, but if the object covers a vertical distance, gravity will occur and its equations are given as follows:

\[ v_f = v_i + at….\text{horizontal motion}\implies \space v_f = v_i + gt….\text{vertical motion} \]

\[ S = v_it + \dfrac{1}{2}at^2….\text{horizontal motion}\implies \space H = v_it + \dfrac{1}{2}gt^2….\text{vertical motion} \]

\[ 2aS = v^{2}_{f} – v^{2}_{i}….\text{horizontal motion}\implies \space 2gS = v^{2}_{f} – v^{2}_{i}….\text{vertical motion} \]

Where $H$ is the height of the object from the ground, $g$ is the gravitational acceleration acting on the object, and its value is $9.8 m/s^2$.

Expert Answer

We are given the following information:

  1. The initial velocity is with which the rock is thrown $v_i = 15\space m/s$,
  2. The time it takes for the rock to reach back $t = 20\space s$,
  3. The initial location of the rock $x = 0$.

Now we are going to take help from the second equation of motion under gravity:

\[ x = v_it + \dfrac{1}{2}gt^2\]

Plugging in the values:

\[ 0 = 15\times 20 + \dfrac{1}{2}(a)(20)^2\]

\[ 15\times 20 = -\dfrac{1}{2}(400a)\]

\[ 300 = -200a \]

\[ a = -\dfrac{300}{200} \]

\[ a = -1.5\space m/s^2 \]

Therefore, the acceleration is of magnitude $1.5\space m/s^2$ and the negative sign indicates that the direction of motion is downward.

Numerical Result

The acceleration comes out to be of magnitude $1.5\space m/s^2$ and the negative sign here indicates that the direction of motion is downward.


The player kicks the football $25.0m$ from the goal, with the crossbar $8.0m$ high. The speed of the ball is $20.0 m/s$ when it leaves the ground at an angle of $48^{\circ}$ horizontally, how long does the ball stay in the air before reaching the goal area? How far does the ball land from the crossbar? And does the ball reach the crossbar while going up or falling down?

Since the ball is moving in the horizontal direction, the velocity component would look like this:

\[v_{0x} = v_0\cos \theta \]

And the distance formula:

\[\bigtriangleup x = v_{0x} t\]


\[t= \dfrac{\bigtriangleup x}{v_{0x}}\]

\[t= \dfrac{25.0 m}{20.0 \cos (48)}\]

\[t= 1.87\space s\]

To find the vertical distance of the ball:

\[y=v_0\sin\theta t – \dfrac{1}{2}gt^2\]

\[y=20\sin (48) (1.87) – \dfrac{1}{2}(9.8)(1.87)^2\]

\[y=10.7\space m\]

Since the ball is $10.7m$ high, it clears the crossbar by:


To find the rise or fall of the ball while it approaches the crossbar:

\[v_y=v_0y – gt\]

\[v_y=v_0\sin\theta – gt\]

\[v_y=20\sin(48) – (9.8)1.87\]

\[v_y=-3.46\space m/s\]

The negative sign tells that it is falling.

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