Question

A stone is dropped from the top of a building strikes the ground with a velocity of 30m/s calculate the height of the building and time taken to reach the ground.


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Answer 1

Explanation:

Initial velocity of the rock u=0

Initial velocity of the rock u=0Final velocity of the rock when it reaches the ground v=30m/s

Initial velocity of the rock u=0Final velocity of the rock when it reaches the ground v=30m/sUsing v

Initial velocity of the rock u=0Final velocity of the rock when it reaches the ground v=30m/sUsing v 2

Initial velocity of the rock u=0Final velocity of the rock when it reaches the ground v=30m/sUsing v 2 −u

Initial velocity of the rock u=0Final velocity of the rock when it reaches the ground v=30m/sUsing v 2 −u 2

Initial velocity of the rock u=0Final velocity of the rock when it reaches the ground v=30m/sUsing v 2 −u 2 =2aS where a=g=10m/s

Initial velocity of the rock u=0Final velocity of the rock when it reaches the ground v=30m/sUsing v 2 −u 2 =2aS where a=g=10m/s 2

Initial velocity of the rock u=0Final velocity of the rock when it reaches the ground v=30m/sUsing v 2 −u 2 =2aS where a=g=10m/s 2

Initial velocity of the rock u=0Final velocity of the rock when it reaches the ground v=30m/sUsing v 2 −u 2 =2aS where a=g=10m/s 2 ∴ 30

Initial velocity of the rock u=0Final velocity of the rock when it reaches the ground v=30m/sUsing v 2 −u 2 =2aS where a=g=10m/s 2 ∴ 30 2

Initial velocity of the rock u=0Final velocity of the rock when it reaches the ground v=30m/sUsing v 2 −u 2 =2aS where a=g=10m/s 2 ∴ 30 2 −0

Initial velocity of the rock u=0Final velocity of the rock when it reaches the ground v=30m/sUsing v 2 −u 2 =2aS where a=g=10m/s 2 ∴ 30 2 −0 2

Initial velocity of the rock u=0Final velocity of the rock when it reaches the ground v=30m/sUsing v 2 −u 2 =2aS where a=g=10m/s 2 ∴ 30 2 −0 2 =2(10)H ⟹H=45 m

Answer 2

Initial velocity of the rock u=0

Final velocity of the rock when it reaches the ground v=30m/s

Using v² −u² =2aS where a=g=10m/s²

∴ 30² −0² =2(10)H ⟹H=45 m

Time :

Now, by using first equation of motion

•v= u + at

substitute the value we get

⇒0 = 30 + 9.8 × t

⇒t = 30/9.8

⇒t = 3.06 seconds