Question

A ball is thrown vertically upwards. It goes to height 19.6 m and then returns back to the ground, find-

1. The initial velocity of the ball.
2. The total time taken of journey.
3. The final velocity of the ball when it strikes the ground (g = $9.8m{s}^{-2}$)

Answer 1

ATQ, height, $s=19.6m$

(i) Initial velocity, $u=?$

Now, when thrown vertically upwards,

Acceleration= $-g=-9.8m/s^2$

Also, at end point, final velocity= $0m/s$

$\Rightarrow v^2=u^2+2as$

So, $(0{)}^2=(u{)}^2+2(-9.8)\left(19.6\right)$

$\Rightarrow u=19.6m/s$

(ii) For going upward,

$v=u+at$

$\Rightarrow 0=19.6+(-9.8)\left(t\right)$

$\Rightarrow t=2s$

Total time= upward+ downward

$\Rightarrow 2+2=4s$

(iii) For downward motion,

$u=0m/s$

$v=?$

$a=+9.8m/s$

$t=2s$

$\Rightarrow v=u+at$

$=19.6m/s$