Question

a stone is thrown in vertically upward direction with a velocity of 5 m/s. is the acceleration of the stone during its upward motion be 10m/s square. What will be the height attained by the stone how much time will it take to reach here​

Answer 1

Answer:

• The Height attained (H) is 1.25 meters
• The time taken (T) to reach height is 0.5 second

Given:

1. Initial velocity (u) = 5 m/s
2. Acceleration (a) = 10 m/s²

Explanation:

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Firstly, as we know the ball is moving up so the acceleration of the stone should be negative i.e. - 10 m/s² as the stone is moving against the gravitational pull. and, the final velocity (v) will be zero as at the highest point the velocity of any object is zero.

Now, applying third kinematic equation,

⇒ v² - u² = 2 a s

Here,

• S Denotes Displacement (Here Height)
• u Denotes Initial velocity.
• v denotes final velocity.
• a denotes acceleration.

Substituting the values,

⇒ (0)² - (5)² = 2 × - 10 × H

⇒ 0 - 25 = - 20 × H

⇒ - 25 = -20 H

⇒ 20 H = 25

⇒ H = 25 / 20

⇒ H = 1.25

⇒ H = 1.25 m

∴ The Height attained (H) is 1.25 meters.

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For time period we need to apply first kinematic equation,

⇒ v = u + a t

Here,

• v denotes final velocity.
• u Denotes Initial velocity.
• t denotes time taken.
• a denotes acceleration

Substituting the values,

⇒ 0 = 5 + (- 10) × t

⇒ - 5 = - 10 × t

⇒ - 5 = - 10 t

⇒ 10 t = 5

⇒ t = 5 / 10

⇒ t = 1 / 2 = 0.5

⇒ t = 0.5 Sec

∴ The time taken (t) to reach height is 0.5 second.

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Answer 2

u, = 5

a, = -10

v, = 0

By using

v= u +at

0 = 5+(-10)T

-5 = -10 T

t = 1/2=0.5

s=ut+1/2at ^2

s = 5(1/2) + 1/2(-10) ×1/4

s = (5/2-5/4) m

s = (10-5 / 4) m

s = 5/4 m

answer = s, = 1.25m

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