A ball is launched horizontally off the top of a building?

Q: A ball is launched horizontally off the top of a building of height H. The intitial velocity is v0 î.

a) Express the horizontal distance D from the bottom of the building that the ball hits the ground at in terms of v0, H, and g

c) What is the speed of the ball the moment it hits the ground in terms of v0, H, and g.

A: (a) Since the initial vertical velocity is zero,

H = ½gt² gives us

t = √(2H / g)

so the horizontal distance

D = v0 * t = v0 * √(2H / g) ◄

c) Vy = g*t = g√(2H/g) = √(2Hg)

so

speed = √(v0² + Vy²) = √(v0² + 2Hg) ◄

Hope this helps!

Let initial horizontal velocity = Vxo = Vo and vertical velocity = Vy(t) = g*t since Vyo = Vy(0) = 0

How long does it take to fall H? h(t) = H + Vyo*t - ½*g*t² = H + 0*t - ½*g*t² = 0 at impact. Thus 0 = H - ½*g*t² -----> H = ½*g*t²

t = √2H/g)

The ball travels horizontally Vxo*t = Vo*t while falling at acceleration g

D = Vxo*√2H/g) = Vo*√2H/g) <-----a)

At impact Vx(t) = Vxo = Vo

Vy(t) = g*t = √2Hg)

Vf = √Vo² + 2Hg) <-----c)