Question

A ball is thrown vertically upwards from ground. It goes to a height 20 m and then returns to the ground. Taking acceleration due to gravity g to be 10 ms2, find total time of journey of ball.

Answer 1

Since, When body goes upward then its Final Velocity becomes Zero and when it returns back to ground then its Initial Velocity becomes Zero.

Velocity for Upward motion.

=> v^2 = u^2 - 2gS

=> 0^2 = u^2 - 2 x 10 x 20

=> 400 = u^2

$= > u \: \: = \sqrt{400} = 20 \: m {s}^{ - 1}$

Now,

Velocity For Downward motion,

=> v^2 = u^2 + 2gS

=> v^2 = 0 + 2 x 10 x 20

=> v^2 = 400

$= > v \: \: = \sqrt{400} = 20 \: m{s}^{ - 1}$

Now,

Time for upward motion,

$=> v = u - gt_{1}$

$=> 0 = 20 - 10t_{1}$

$=> 10t_{1} = 20$

$=> t_{1} = \dfrac{20}{10} \: s$

$=> t_{1} = 2 \: s$

Again,

$=> v = u + gt_{2}$

$=> 20 = 0 + 10 \times t_{2}$

$=> \dfrac{20}{10} = t_{2}$

$=> t_{2} = 2\: s$

Now,

$=> Total\: time = t_{1}+ t_{2}$

$=> T = (2 + 2) s$

$=> T = 4 \: s$