# A Girl Throws A Ball Vertically Upwards (Linear Motion)

A girl throws a ball vertically upwards with an initial velocity of 20m/s and then catches it sometime later. Calculate the time taken for the ball to return to the same height it was released from!

SOLUTION

We will take anything moving upwards as positive in this case.

The distance travelled in the positive and negative direction cancels out because the ball returns to its original position. Hence, the displacement is zero.

The final velocity is the velocity at which the girl catches the ball. Since the girl catches the ball at the same height (and provided the air has a negligible effect on the ball), the final velocity will be -20m/s (upwards direction positive, downwards direction negative).

For the acceleration, when the ball is tossed upwards, it decelerates due to the gravitational pull, but because the upwards direction is taken as positive, the ball decelerates in the positive direction. As the ball reaches its maximum height and moves downwards, it accelerates in the negative direction. So, when moving down, the acceleration will be -9.81m/s2, which is the constant for gravitational acceleration.

Let’s use the first linear equation of motion: v = u+at

v = -20 m/s (downwards direction negative)

u = 20 m/s (upwards direction positive)

g = -9.81 m/s2

t = …?

-20 = 20 + (-9.81)t

9.81 t = 40

t = 40/9.81

t = 4.08 s 