What height (displacement) will a ball reach if thrown upward with an initial velocity of 15 m/s

11.48 m

Explanation:

initial velocity (u) = 15 m/s

acceleration due to gravity (g) = 9.8 m/s

final velocity (v) = 0 (the ball's speed reduces gradually when thrown until

becomes 0 at maximum height)

find the height reached.

height (s) = ut + 0.5 at^{2}

we cannot apply this formula directly because we do no know the time

(t), so we first need to find the time

v = u + at

0 = 15 + (-9.8 x t) (the negative sign is because the ball is decelerating

upwards)

15 = 9.8t

t = 1.53 s

now that we have the time we can put it into the initial equation height (s) = ut + 0.5 at^{2}

height (s) = (15 x 1.53) + (0.5 x (-9.8) x 1.53^{2})

height = 22.95 - 11.47 = 11.48 m