Question

a ball is thrown vertically upwards from the ground . it crosses a point at the height of 25meter twice at an interval of 4 seconds at what velocity the ball was thrown

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Answer 1

Answer:

29.5 m/s

Explanation:

You're throwing a ball vertically upwards, so that means that it'll eventually fall down after reaching a certain height. It says that the ball crosses a point at the height of 25 meters twice i.e one during the upward motion and other during downward motion.

Interval between them = 4 seconds.

Interval between initial crossing and at maximum height = 2 seconds.

At the first crossing,

u = ?

v = 0 m/s

a = -9.8 m/s^2

t = 2 s

v = u + at

=> u = v - at

=> u = 0 + 9.8 x 2 = 19.6 m/s

If we consider from the ground,

u = ?

v = 19.6 m/s

s = 25 m

a = -9.8 m/s^2

v^2 - u^2 = 2as

=> u^2 = v^2 - 2as

=> u^2 = 19.6^2 + 2x9.8x25

=> u^2 = 874.16

=> u = 29.5 m/s

Initial velocity is 29.5 m/s