Question

A stone is thrown vertically upward with anniniial velocity os 20m/s find the time taken by it to reach the height -(g -10m/s
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Answer 1

Given:-

Final velocity ,v = 0m/s

Initial velocity ,u = 20m/s

Acceleration due to gravity ,g = 9.8m/s²

To Find:-

Maximum height ,h

Time taken ,t

Solution:-

Firstly we calculate the maximum height attained by the stone.

Using 3rd Equation of Motion

• v² = u² +2ah

v is the final velocity

g is the acceleration due to gravity

u is the initial velocity

h is the height attained by stone

Substitute the value we get

→ 0² = 20² + 2× (-9.8) × h

→ 0 = 400 + (-19.6 ) × h

→ -400 = -19.6×h

→ h = -400/-19.6

→ h = 400/19.6

→ h = 20.40 ≈ 20 m

Therefore, maximum height attained by the stone is 20 metres.

Now, Calculating the time taken

Using 1st Equation of Motion

• v = u +at

Substitute the value we get

→ 0 = 20 + (-9.8) ×t

→ -20 = -9.8×t

→ t = -20/-9.8

→ t = 20/9.8

→ t = 2.04 s

Therefore,. the time taken by the stone to reach maximum height is 2.04 second .