Question

a ball is thrown up with 2 speed of 15m/s.how high will it go before it begans to fall?take g=9.8m/s2​

Answer 1

By using equation of motion

S = u² / (2g) ………[∵ Velocity at maximum height is 0]

= (15 m/s)² / (2 × 9.8 m/s²)

= 11.48 m

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.