Question

A ball is dropped from rest. After time t the ball has travelled a distance of x. After time 5t the ball has travelled a distance of?

Answer 1

Given:

Distance travelled after time t = x

Given that the ball was dropped from rest, let's find the distance the ball travelled after time 5t.

To find the distance, apply the motion formula:

$h=\frac{1}{2}*gt^2$

Where:

g is the acceleration due to gravity = 9.8 m/s²

h is the vertical distance travelled.

t is the time.

After time t, we have:

$x=\frac{1}{2}gt^2$

After time 5t, we have:

$h=\frac{1}{2}g(5t)^2$

Now divide the second expression by the first expression.

$\begin{gathered} \frac{\frac{1}{2}g(5t)^2}{\frac{1}{2}gt^2} \\ \\ \frac{(5t)^2}{t^2} \\ \\ =\frac{5^2t^2}{t^2} \\ \\ =\frac{25t^2}{t^2} \\ \\ =25 \end{gathered}$

Therefore, after the time 5t, the ball has travelled a distance of 25x.

• ANSWER:

25x