Question

A ball is thrown up with the speed of 15 metre per second. How high will it go before it beings to fall? (The value of g = 9.8 M per second square)

Answer 1

Final speed at the top most point (v) = 0

Initial speed (u) = 15 m/s

a = -g = -9.8

Hence, as per the third law of motion,/

v square = u square + 2.a.s

or distance s = (v square - u square)/2.a

or s = (0 - 225)/(2 x -9.8)

or s = 225/19.6 = 11.48 mtrs Answer




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Answer 2

Solutions:

Please note that here the ball is going up against the gravity, so the value of g is to be taken as negative.

Here, initial speed, u = 15 m/s

Final speed, v = 0

Acceleration due to gravity, g = -9.8 m/s^2

And, height, h = ?

Now, putting all these values in the formula; v^2 = u^2 + 2gh, we get;

=> (0)^2 = (15)^2 + 2 × (-9.8) × h

=> 0 = 225 - 19.6h

=> 19.6h = 225

=> h = 225/19.6

=> h = 11.4 m.

Thus, the ball will go to the maximum height of 11.4 metres before it begins to fall.