Question

A body is released from the top of a power of beight it takes t see to reach the ground. Where will
be the ball after time 1/2 sec
2) At W2 from the ground
b) At 4 from the ground
Depends upon mass and volume of the body d) A 34 from the ground
het If the initial velocity is 21 the
AD​

Answer 1

Answer:

Given:-

u = 0 m/s

t = 1/2 s

a = g = 10 m/s²

To find:-

The position of the ball after 1/2 s

Formula to be used:-

S = ut + (1/2)*a*t²

How to find:-

Applying the formula S = ut + (1/2)*a*t²

S = 0*(1/2) + (1/2)*10*(1/2)²

S = 0 + 5(1/2)²

S = 5/4 m

Hence the ball will be at a distance of 1.25 m from the top of the tower.

If the height of the tower is given then the distance of the ball from the ground at 0.5 s is (h - 1.25) m. Where the height of the tower is 'h'.

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