Question

A ball is thrown vertically upward from the ground with a speed of 15m/s. On the way down, it is caught at a point 6m above the ground. How long did the trip take?

Answer 1

To determine the time it took for the ball to complete its trip, we can analyze its motion separately for the upward and downward phases.

Let's start by considering the upward phase of the ball's motion. The initial velocity ($u$) is 15 m/s, and we know that the final velocity ($v$) at the peak of the ball's trajectory will be zero, as it momentarily comes to a stop before falling back down. The acceleration ($a$) acting on the ball throughout its motion is due to gravity and is approximately equal to -9.8 m/s² (negative because it acts in the opposite direction of the motion).

We can use the following equation of motion to find the time ($t_1$) it takes for the ball to reach the peak:

$v=u+at$

Since the final velocity ($v$) at the peak is 0, the equation becomes:

$0=15+(-9.8)t_1$

Solving for $t_1$, we have:

$9.8t_1=15$

$t_1=\frac{15}{9.8}$

Now, let's consider the downward phase of the ball's motion. The initial velocity is zero at the peak, and the final velocity ($v$) when it reaches a height of 6 m above the ground is unknown. The acceleration (aa<span class="strut" style="height: 0.4306e