Vertical Ball Motion: Distance and Displacement in 3 Seconds

Question:

A ball is thrown vertically upward with a velocity of 20 ms^-1.Find

• (a) The distance travelled by the ball in first three seconds.
• (b) Displacement of the ball in 3 seconds. (take g=10 ms^-2 )

Solution:

Thrown up- Sign convention

(a) Distance travelled by the ball in first three seconds

When a ball is thrown vertically upward, the velocity at its highest point is zero (v=0)

Here initial velocity of the ball (u) =20 ms^-1

Hence the time taken to reach the maximum height is $${t=\frac{u}{g}=\frac {20}{10} = 2 \ sec}$$

Distance travelled by the ball in 3sec is = (Distance travelled from t=0 s to t=2 s+Distance travelled from 2s to 3s}

$${s=(ut-\frac{1}{2}gt^2) +\frac{1}{2}gt^2}$$

$${s=[20(2)-\frac{1}{2}10(2^2)]+\frac{1}{2}10(1^2)}$$

$${s=[40-20] +5}$$

$$s= 25 \ m$$

(b) Displacement of the ball in 3 seconds

$${Displacement = 20 \ m – 5 \ m=15 \ m}$$

The displacement of the ball in 3 seconds is 15 m