Question #a3917

1 Answer
Jan 25, 2017

Initial upward velocity of the stone thrown $= u = 29.8 m \text{/s}$

Retardation acting on the stone due to downward gravitational force is equal to the acceleration due to gravity.
So $g = - 9.8 m \text{/} {s}^{2}$

i) If stone attains velocity $v$ after $t$ sec then by equation of kinematics we have

$v = u - g t$

At maximum height the velicity of the stone will be zero.
So $0 = 29.8 - 9.8 \times t$

$\implies t = \frac{29.8}{9.8} \approx 3.04 s$

ii) Again ${v}^{2} = {u}^{2} - 2 g h$,where h is height attained at $t$ sec

At maximum height $v = 0 \mathmr{and} h = {h}_{\text{max}}$

So ${h}_{\text{max}} = {u}^{2} / \left(2 g\right) = {29.8}^{2} / \left(2 \times 9.8\right) \approx 45.3 m$

iii) And the velocity after 3 sec

${v}_{3} = 29.8 - 9.8 \cdot 3 = 0.4 m \text{/s}$