Question

On an alien planet, a ball is thrown vertically upwards from the ground, rises to a height of 100 m, and then falls back to the ground. The total time is taken from when the ball was thrown upwards to when it reached the ground again, is 10 seconds.....?

....What is the acceleration due to gravity on that planet?

Answer 1

`g=8"m"//"s"^2`

Explanation:

We can find the gravitational acceleration on the planet using kinematics. We will first make the assumption that air resistance and all other outside forces are negligible and consider only the effects of the force of gravity.

We have the following information:

• `v_i=0" m"//"s"`
• `Deltay=100"m"`
• `Deltat_"total"=10"s"`

We will use the following kinematic equation and solve for `a`:

`color(blue)(Deltay=v_"iy"Deltat+1/2a_y(Deltat)^2),`

where `v_"iy"` is the initial vertical velocity, which we know is zero as the ball begins from rest (we would usually use an initial velocity and set the final velocity, i.e. that at the maximum altitude, to zero, but without a known acceleration we cannot solve for velocity, so we will make this assumption)

Because the launch and landing altitudes are the same for the ball (the ground), we can use the fact that the rise time and the fall time are also the same. This means that the elapsed time for the ball to reach its maximum altitude after launch is `(10"s")/2=5"s"`. We are given the maximum altitude of the ball, so, solving for a:

`color(blue)(a=(2Deltay)/(Deltat)^2)`

Using our known values:

`a=(2*100"m")/(5"s")^2,`

`=(200"m")/(25"s^2)`

`=8"m"//"s"^2`

Therefore, the gravitational acceleration on this planet is `8"m"//"s"^2`, about 80% of that on Earth.