A stone thrown vertically upwards attains a maximum height of 45 m. In what time the velocity of stone become equal to one-half the velocity of throw? (. Given g=10 ms -2 )

Question

A stone thrown vertically upwards attains a maximum height of 45 m. In what time the velocity of stone become equal to one-half the velocity of throw? (. Given g=10 ms -2 ) (A) 2 s (B) 1.5 s (C) 1 s (D) 0.5 s

Tardigrade Admin · Accepted Answer

Let us solve the problem in terms of relative initial velocity, relative acceleration and relative displacement of the coin with respect to floor of the lift.
Given,

u = 0 m s^{- 1}, a = 9.8 m s^{- 2}, s = 4.9 m, t = ?

As,

4.9 = 0 \times t + \frac{1}{2} \times 9.8 \times t^{2}

Or

4.9 t^{2} = 4.9

Or

t = 1 s

or

15 = 30 - 10 t

10 t = 15

Or

t = 1.5 s

Q. A stone thrown vertically upwards attains a maximum height of 45m. In what time the velocity of stone become equal to one-half the velocity of throw? ( Given g=10ms−2 )

2 s

1.5 s

1 s

0.5 s

Let us solve the problem in terms of relative initial velocity, relative acceleration and relative displacement of the coin with respect to floor of the lift. Given, u=0ms−1,a=9.8ms−2,s=4.9m,t=? As, 4.9=0×t+21​×9.8×t2 Or 4.9t2=4.9 Or t=1s or 15=30−10t 10t=15 Or t=1.5s

Q. A stone thrown vertically upwards attains a maximum height of $45 m$ . In what time the velocity of stone become equal to one-half the velocity of throw? $($ Given $g = 10 m s^{- 2}$ )