Question

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

Answer 1

Answer:

The ball will go 11.48 m high before it begins to fall

Explanation:

Final velocity, v = u + at

When it begins to fall, v = 0

Initial velocity, u = 15 m/s

Acceleration a = -9.8 m/s²

0 = 15 + -9.8*t

t = 15/9.8 = 1.53 s

Distance, S = ut + 0.5 at² = (15*1.53) + (0.5*-9.8*1.53²) = 22.95 - 11.47 = 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.