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July 30, 2014

July 30, 2014

Posted by **201** on Thursday, October 6, 2011 at 8:21pm.

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?

- Physics -
**Steve**, Friday, October 7, 2011 at 2:07pmThe 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

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