Question

In an action movie the lead character must jump off a balcony at an angle of 20 degrees, 10 meters above the ground and land in a window 3 meters away and 8.5 metters abover the ground.

What should the velocity be in order to make this jump?

How long is the stunt double in the air?

What is the velocity as the stunt double impacts the window?

What is the maximum height the double is above the ground?

What is the acceleration at the highest point of the jump?

Answer 1

The initial velocity has x- and y-components:

Vx = V cos 20 = 0.9397V
Vy = V sin 20 = 0.3420V

How long does it take to cross the 3m?

t = 3/Vx = 3/.9379V = 3.1986/V seconds
V = 3.1986/t

How long does it take to fall 1.5m to the window?

Py = 10 + .3420Vt - 4.9t^2 = 8.5
10 + .3420t*3.1986/t - 4.9t^2 = 8.5
4.9t^2 - 1.0939 - 1.5 = 0
4.9t^2 - 2.5939 = 0
t = sqrt(2.5939/4.9) = 0.7275 sec

V = 3.1986/.7275 = 4.3967 m/s

Check to make sure of height:
Py = 10 + .3420*4.3967*.7275 - 4.9*.5293 = 8.500 m

So, velocity = 4.4m/s
Time in air = 0.73 s
V at window = sqrt(Vx + Vy)^2 = sqrt(4.13^2 + 2.06^2) = 4.61 m/s

Acceleration never changes: -9.8m/s^2