Stunt Car Jump: How Far and How Fast? | Solving for Distance and Impact Speed"

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A stuntman drives a car off a 30-meter-high cliff at a speed of 20 m/s, with the road inclined at 20 degrees. The horizontal and vertical components of the initial velocity are calculated as 18.8 m/s and 6.84 m/s, respectively. To determine how far from the cliff the car lands, the time of flight is derived from the vertical motion equation, leading to a quadratic equation for time. The horizontal distance is then calculated using the time found. The impact speed is determined using the final velocity components from both the horizontal and vertical motions.
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Homework Statement



A stunt man drives a car at a speed of 20 m/s off a 30-m-high cliff.
The road leading to the cliff is inclined upward at an angle of 20(degrees).
a. How far from the base of the cliff does the car land?
b. What is the car's impact speed?

Homework Equations





The Attempt at a Solution



Here is my solving:

Vix=Vicos@
=20cos20
=18.8m/s

Viy=Visin@
=20sin20
=6.84m/s

a) How far from the base of the cliff does the car land?
x=Vix t
x=18.8t ------(1)

y=Viy t - 1/2gt^2
-30=6.84t-0.5(9.8)t^2
-4.9t^2+6.84t+30=0 (divide this by -4.9)
t^2-1.39t+30=0
From this equation we can find t then sustitute its magnitude in equation 1, so we can find x

b)speed=sqrt((Vfx)^2+(Vfy)^2)
 
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