Question

A ball with mass m kg is thrown upward with initial velocity 19 m/s from the roof of a building 12m high. Neglect air resistance. Use g =9.8 m/s^2. Find the maximum height above the ground that the ball reaches.

Answer 1

the max. height formula is
\(\large h_{max} = \dfrac{u^2 \sin^2 \theta }{2g}\)

here, since the ball is thrown vertically upwards,
\(\theta = 90, \quad \sin 90 =1\)
and since you're already on a 12 m height building,
total height from ground H will be
\(\large H= 12 +\dfrac{u^2}{2g}\)

u =19
g=9.8
just plug in!

Answer 2

Thank you @hartnn!!

Answer 3

o hey, you're welcome ^_^