Question

A projectile is fired at 25m/s on top of a roof 40m high at an angle of 60°. At what speed does it land on the ground ?

Answer 1

how long does it take to hit?
40+25sin60°t-4.9t^2 = 0
t = 5.82s

The velocity has two components:
vy = 25sin60°-9.8t = 25*0.866-9.8*5.82 = -35.386
vx is constant at 25cos60° = 12.5
landing speed is thus
s^2 = 12.5^2 + 35.386^2 = 1408.42
s = 37.53 m/s

Answer 2

Vo = 25m/s[60o].
Xo = 25*Cos60 = 12.5 m/s = Hor. component of initial velocity.
Yo = 25*sin60 = 21.7 m/s = Ver. component of initial velocity.

Y^2 = Yo^2 + 2g*h = 0,
21.7^2 + (-19.6h) = 0,
h = 24 m. above the bldg.

ho = 40 + 24 = 64 m. above gnd.

Y^2 = Yo^2 + 2g*ho = 0 + 19.6*64 = 1254,
Y = 35.4 m/s = Ver. component of final velocity.

V = sqrt(Xo^2 + Y^2) = sqrt(12.5^2 + 35.4^2) = 37.5 m/s.