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

Given,

$V e l o c i t y = 5 m / s, a = 10 m / s^{2}$

So, the height is:

$v^{2} - u^{2} = 2 a s$

$s = \frac{v^{2} - u^{2}}{2 a}$

$= \frac{0 - 25}{2 \times (- 10)} = 1.25$ since $v = 0$

Since $a$ is negative so its direction is opposite to the direction of motion.

So,

$t = \frac{v - u}{a} = \frac{0 - 5}{- 10} = \frac{1}{2}$

$t i m e = 0.5 s$

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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 10 ms−2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Given,Velocity=5m/s,a=10m/s2So, the height is:v2−u2=2ass=v2−u22a=0−252×(−10)=1.25 since v=0Since a is negative so its direction is opposite to the direction of motion.So, t=v−ua=0−5−10=12time=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?