Question

a stone is thrown in a vertically upward direction with a velocity of 10m/s if the acceleration of the stone during its motion is 10m/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?

Answer 1

Answer:

V^2 = U^2 - 2ah

U^2 = 2ah

h = 100 / 2*10 = 5m

V= U - at

t = U/g = 10/10 = 1sec

Answer 2

Answer

• Height , s = 5 m
• Time , t = 1 s

Given

• A stone is thrown in a vertically upward direction with a velocity of 10 m/s if the acceleration of the stone during its motion is 10 m/s² in the downward direction

To Find

• Height attained by the stone
• Time taken to reach the height

Concept Used

Equation's of motion

• v = u + at
• v² - u² = 2as

Solution

Initial velocity , u = 10 m/s

Acceleration due to gravity , a = g = - 10 m/s²

[ ∵ thrown against the gravity ]

Final velocity , v = 0 m/s

[ ∵ At extreme position stone goes to rest ]

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Apply 3 rd equation of motion ,

⇒ v² - u² = 2as

⇒ (0)² - (10)² = 2(-10)s

⇒ 0 - 100 = -20s

⇒ - 100 = - 20s

⇒ 100 = 20s

⇒ 20s = 100

⇒ 2s = 10

⇒ s = 5 m

So , maximum height attained by stone , s = 5 m

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Apply 1 st equation of motion ,

⇒ v = u + at

⇒ (0) = (10) + (-10)t

⇒ 0 = 10 - 10t

⇒ 10t = 10

⇒ t = 1 s

So , stone takes 1 second to attain it's extreme position

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