Question

A stunt man drives a car at a speed of 20 m/s off a 20-m-high cliff. The road leading to the cliff is inclined upward at an angle of 20∘.

Answer 1

car
O(launch speed (u) at angle 20 with ox
o____________ x
|....
|(h)....
|.............
|................. O lands in curve parabola

vertical speed (down) = u sin 20
hori speed = u cos 20 ....... (same always)
y = u sin 20 *t + 0.5 g t^2 = 30
x = u cos 20 * t = R (range)
here (t) time of flight > eliminate
20 sin 20 *t + 0.5*9.8 t^2 = 30
6.84 t + 4.9 t^2 - 30 = 0
it gives >> t = + 1.873 s (leaving other -ve time)
R = range where lands = 20* cos 20 * 1.873
R = 35.2 meter from base
============================
impact speed
energy conservation>
initial = final
0.5 mu^2 + mgh = 0.5 mv^2 + 0 (at base)
u^2 + 2gh = v^2
v^2 = 20*20 + 2*9.8*30 = 400 + 588 = 988
v = impact speed = 31.43 m/s

other method>>>

vy = dy/dt = u sin 20 + g t = 25.195
vx = u cos 20 = 18.794
v^2 = [vx^2+ vy^2]^1/2 = 31.43 m/s

Answer 2

How far from the base of the cliff does the car land?