A stone is thrown in a vertically upward direction with a velocity of $5 m s^{- 1}$ . If the acceleration of the stone during its motion is $10 m s^{- 2}$ in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

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Solution

Step 1: Given
Initial velocity (u) = $5 m s^{- 1}$
Acceleration = $- 10 m s^{- 2}$
Final velocity (v) = $0$
Step 2: Formula and calculation
$v^{2} = u^{2} + 2 a s 0 = 5^{2} + 2 (- 10) \times s 20 s = 25 s = \frac{25}{20} s = 1.25 m$
Hence, the maximum height is $1.25 m$ .
Now,
$v = u + a t 0 = 5 + (- 10) t 10 t = 5 t = \frac{5}{10} t = 0.5 s$
Hence, time taken to reach maximum height is $0.5 s$ .

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A stone is thrown in a vertically upward direction with a velocity of 5ms-1. If the acceleration of the stone during its motion is 10ms-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Step 1: GivenInitial velocity (u) = 5ms-1Acceleration = -10ms-2Final velocity (v) = 0Step 2: Formula and calculationv2=u2+2as0=52+2(-10)×s20s=25s=2520s=1.25mHence, the maximum height is 1.25m.Now,v=u+at0=5+(-10)t10t=5t=510t=0.5sHence, time taken to reach maximum height is 0.5s.

A stone is thrown in a vertically upward direction with a velocity of $5 m s^{- 1}$ . If the acceleration of the stone during its motion is $10 m s^{- 2}$ in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?