Question

A body drops a ball from the top
af a
tower of height 19.6m .
calculate the velocity of the ball
just before lt touches the
ground? ​

Answer 1

Initial velocity u = 0

Final velocity v = ?

Height s = 19.6 m

From III equation of motion,

v² = u² + 2gs

v² = 0 + 2 × 9.8 × 19.6

v² = 384.16

v = √384.16

v = 19.6 m/s

So, the velocity of the ball just before it touches the ground when dropped from the top of a tower of height 19.6 m is 19.6 m/s.

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Answer 2

V2 - U2 = 2 gs.....

Maybe 19.6 m/s