Question

3.A ball is thrown up with a speed of 15 m/s. How high will it go before it begins to fall? Take g=9.8 m/s^2.

Answer 1

u = 15m/s

g = 9.8

use v^2 = u^2 + 2as

we need to find out the height of the ball before it falls down.

we know that final velocity is 0 at the highest point because the ball comes to rest and then falls down.

hence 0 = 225 + 2 x 9.8 x s

-225/19.6 = s

s = 11.47 or 11.5 m.

We are taking mod value because height is not negative.

if you take g = 10 m/s, you'll get a whole number answer.