Stunt Car

A stunt man drives a car at a speed of 20m/s...?

A stunt man drives a car at a speed of 20m/s off a 30-m-high cliff. The road leading to the cliff is inclined upward at an angle of 20 degree.

A. How far from the base of the cliff does the car land?
B. What is the car's impact speed?

Make two charts, one of "x"s and one of"y"s, both with time (t), initial velocity, final velocity, the change in x or y, and acceleration(a) )(acceleration in the x-direction is always 0 and acceleration in the y direction is always -9.8 m/s^2, the force of gravity)

To find the initial velocities in the x and y directions, take the x and y components of the right triangle created with an angle of 20 degrees and a hypotenuse of 20 (m/s).

Use equations such as ∆x=(.5)(v(initial)+v(final)t,
∆x=v(initial)t+.5a(t^2)
and v(final)^2=v(initial)^2+2a∆x to solve for unknown variables.

At the end, use the final velocities in the x and y direction to find the the impact of the car. the final impact will be: sqare root (x^2+y^2)

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