Question

A ball is thrown up with a speed of 15 m/s. How high will it go before it begins to fall? (g = 9.8 m/s²)

Answer 1

y = 11.48 m

Step-by-step explanation:

u = zero; because the speed reached maximum height

minus cancels each other

`{u}^(2) = {v}^(2) - 2gy \\ 0 = {15}^(2) - 2(9.8)y \\ - 225 = - 19.6y \\ y = ( - 225)/( - 19.6) = 11.48 \: m`