A stone is thrown in vertically upward direction with a velocity of 5m

1.	A stone is thrown in 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 he the height attained by the stone and how much time will it take to reach there?
Answer» Initial velocity of a stone, u = 5 ms^-1. Final velocity of a stone v = 0 (at rest) If the acceleration of the stone during its motion is = -10ms^-2. Let the maximum height be ‘s’ v = u + at 0 = 5 + (-10)t t = \(\frac{-5}{10}\) = 0.5s time taken by stone = 0.5s As per 3rd Law of motion v² = u² + 2as (0)² = (5)² + 2(-10)s ∴ S = \(\frac{5^2}{20}=\frac{25}{20}\) = 1.25 m Height attained by the store = 1.25 m.

A stone is thrown in 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 he the height attained by the stone and how much time will it take to reach there?

Answer»

Initial velocity of a stone, u = 5 ms^-1.

Final velocity of a stone v = 0 (at rest)

If the acceleration of the stone during its motion is = -10ms^-2.

Let the maximum height be ‘s’

v = u + at

0 = 5 + (-10)t

t = \(\frac{-5}{10}\) = 0.5s

time taken by stone = 0.5s

As per 3rd Law of motion

v² = u² + 2as

(0)² = (5)² + 2(-10)s

∴ S = \(\frac{5^2}{20}=\frac{25}{20}\) = 1.25 m

Height attained by the store = 1.25 m.