3) A ball is thrown up with a speed of 15 m/s. How high will it go
before it begins to fall?
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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.
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