projectile motion

Let U be the take off speed of the car , and x be the angle with the horizontal at which the car "takes off".

Consider the horizontal motion:
[U.cos(x)].t = 30 ------------------- (1)
where t is the time of flight of the car

Consider the vertical motion:
use equation of motion: v^2 = u^2 + 2.a.s
with v = 0 m/s, u = U.sin(x), a = -g, s = 5 m
hence, 0 = [U.sin(x)]^2 + 2.(-g).(5)
i.e. U.sin(x) = 10 m/s ---------------- (2)
Use equation of motion: v = u + a.t
with v = 0 m/s, u = 10 m/s, a = -g, t = ?
hence, 0 = 10 + (-g).t
i.e. t = 1 s

Therefore, from (1), U.cos(x) = 30 m/s ---------------- (3)
(2)/(3): tan(x) = 10/30
x = 18.43 degrees
hence, U = 30/cos(18.43) m/s = 31.6 m/s

Let v be the take-off velocity of the car and θ be the angle of inclination of the velocity to the horizontal, then

Considering the vertical part of motion, since 5 m is the max. height reached by the car, we have:

0 - (v sin θ)2 = 2(-g)(5)

(v sin θ)2 = 100

v sin θ = 10

Then, the time of flight T can be found by:

(v sin θ) - 10T = 0

T = 1 s

Considering the horizontal part of motion:

(v cos θ)T = 30

v cos θ = 30

So using the Pyth. thm, v = 31.6 m/s which is the take-off speed of the car.