Question

6 A body drops down a tower and reaches the ground in 0.6s. If g = 10 m/s2, then find the height of the tower.
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Answer 1

Answer:

1.8 m

Explanation:

Since the body is dropped

u = 0 m/s

t = 0.6 s

g = 10 m/s^2

You can use the formula

S = ut + 1/2gt^2

S = 0*0.6 + 1/2 * 10 * 0.6 * 0.6

S = 1.8 m

Therefore the height was 1.8 meters

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Answer 2

Answer :

Time taken by body to reach ground = 0.6s

Acceleration due to gravity = 10m/s²

We have to find height of the tower$.$

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◈ For a body falling freely under the action of gravity, g is taken positive.

In free falling motion, acceleration dut to gravity continuously acts in the downward direction and magnitude of g remains constant throughout the motion so we can apply equation of kinematics to solve this question.

Let's apply 2nd eq. of kinematics :

➝ H = ut + (1/2)gt²

➝ H = 0 + (1/2)(10×0.6²)

➝ H = (1/2)(3.6)

➝ H = 1.2m