Question

A ball is thrown vertically upwards from the ground with a velocity 24.5 metre per second calculate the time after which the ball will be at height of 29.4​

Answer 1

Answer :

t = 2 sec and 3 sec

Explanation :

Given,

• A ball is thrown vertically upwards from the ground with a velocity 24.5 m/s

To find,

• the time after which the ball will be at height of 29.4​ m

Solution,

 Let 't' be the time after which the ball will be at height of 29.4 m

➟ initial velocity, u = 24.5 m/s

➟ height, h = 29.4 m

➟ acceleration due to gravity, g = 9.8 m/s²

Second equation of motion,

    $\bf S=ut+\frac{1}{2} at^2$

Here,

• S = h
• a = -g (since the ball is thrown vertically upwards)

 $\bf h=ut-\frac{1}{2} gt^2$

 $\sf 29.4 =24.5t-\frac{1}{2} (9.8)t^2 \\\\ 29.4 =24.5t-4.9t^2 \\\\ 4.9t^2-24.5t+29.4=0 \\\\ 49t^2-245t+294=0 \\\\ 7(7t^2-35t+42)=0 \\\\ 7t^2-35t+42=0 \\\\ 7(t^2-5t+6)=0 \\\\ t^2-5t+6=0 \\\\ t^2-2t-3t+6=0 \\\\ t(t-2)-3(t-2)=0 \\\\ (t-2)(t-3)=0 \\\\ t=2 \ sec \ (or) \ 3 \ sec$

Therefore, the time after which the ball will be at height of 29.4 m is 2 sec and 3 sec