Question

A stone is dropped from the top of tower. if its velocity at the mid point of height of tower is 10m/s then the height of the tower is

Answer 1

Answer:

10 m approximately

Explanation:

Let the height of tower be "H"

Use equation of motion: v² = u² + 2aS

Where

• v = Velocity at point S
• u = Initial velocity
• a = Acceleration
• S = Displacement covered while there is a change in velocity from "u" to "v"

Velocity at mid point is

v = √(u² + 2aS)

10 m/s = √[(0 m/s)² + (2 × 10 m/s² × H/2)]

100 = 10H

H = (100/10) m

H = 10 m

Answer 2

Answer:

lets take the total height as H.

given ,

at H/2 ,

v = 10 m/s

by using the third formula ie v² - u² / 2a = S

10²/ 2×10 = H/2                                               [ as u = 0 ]

100 / 20 = H / 2

5 = H / 2

H = 5×2

   = 10 m

hence , the height of the tower is 10 m.

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