# 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/s^{2}, 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/s^{2}

t = …?

-20 = 20
+ (-9.81)t

9.81 t =
40

t =
40/9.81

**t = 4.08
s**

## Post a Comment for "A Girl Throws A Ball Vertically Upwards (Linear Motion)"