Projectile Motion of Car Problem

triamanda
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Homework Statement



A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24 degrees below the horizontal. The parking break is defective, so the car rolls from rest down the incline with a constant acceleration on 4m/s2 and travels 50m to the edge of the ocean. The cliff is 30 m above the ocean.

a. What is the car's horizontal position relative to the base of the cliff when it lands in the ocean?

b. How long is the car in the air?

Homework Equations



Vy2 = -2gΔy
Δy = -1/2g(Δt)2
Vy= -gΔt
Vx = Δx/Δt

The Attempt at a Solution



Δt = square root (30sin(24)/(1/2)g) = 2.5s

However, I don't really think that's right. If someone could explain the process of solving the overall of equation, that would be extremely helpful.

...finals tomorrow.
 
on Phys.org
triamanda said:

Homework Statement



A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24 degrees below the horizontal. The parking break is defective, so the car rolls from rest down the incline with a constant acceleration on 4m/s2 and travels 50m to the edge of the ocean. The cliff is 30 m above the ocean.

a. What is the car's horizontal position relative to the base of the cliff when it lands in the ocean?

b. How long is the car in the air?


Homework Equations



Vy2 = -2gΔy
Δy = -1/2g(Δt)2
Vy= -gΔt
Vx = Δx/Δt

The Attempt at a Solution



Δt = square root (30sin(24)/(1/2)g) = 2.5s

However, I don't really think that's right. If someone could explain the process of solving the overall of equation, that would be extremely helpful.

...finals tomorrow.

This problem is a bit complicated. First you must calculate the velocity of the car when it reaches the bottom of the incline just before it leaves the cliff. From that velocity vector which makes an angle of 24 degrees below the horizontal, you need to calculate the horizontal and vertical components of its velocity. Call these [itex]V_0x[/itex] and [itex]V_0y[/itex] respectively. Now take your coordinate axes to have their origin at the edge of the cliff where the car leaves. Choose the Y axis vertical, X axis horizontal. Then you need to simultaneously solve the kinematic equations of projectile motion:

[tex]X = X_0 + V_0x t + \frac{1}{2} a_x t^2[/tex]

[tex]Y = Y_0 + V_0y t + \frac{1}{2} a_y t^2[/tex]

Recall that [itex]a_y = g[/itex]. What is [itex]a_x[/itex] ?
 

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