Question

an object is thrown up with a velocity of 19.6 m/s a)find the total time taken by it to return back b) how high will it go c)what is the acceleration after 5 s​

Answer 1

Given:-

• Initial velocity (u) = 19.6m/s

• Final Velocity (v) = 0m/s

• Time Taken (t) = 5s

To Find:-

(a) Total time taken by it to return back.

(b) Height attained by it.

(c) Acceleration after 5s.

Solution:-

We know that , the acceleration due to gravity at the earth surface is 9.8m/s². Acceleration due to gravity always acts in downward direction

∴ We take acceleration in negative = -9.8m/s².

(b)

By using third equation of motion

⟹ v² = u² + 2as

Put the values , we get

⟹ 0² = 19.6² + 2× 9.8 ×s

⟹ 0 = 384.16 - 19.6×s

⟹ - 384.16 = 19.6×s

⟹ s = -384.16/-19.6

⟹ s = 19.6

∴ The height attained by object is 19.6m

(a)

By using 1st equation of motion

⟹ v = u+at

Put the values,we get

⟹ 0 = 19.6 + (-9.8)×t

⟹ -19.6 = -9.8×t

⟹ -19.6/9.8 = t

⟹ t = 2s

∴ The time taken by it to return back is 2s.

(c)

By using 1st equation of motion

⟹ v = u+at

Put the value,we get

⟹ 0 = 19.6 + a×5

⟹ a = -19.6/5

⟹ a = -3.92 m/s²

here negative sign show deceleration

∴ The acceleration of object is 3.92 m/s².