Basic Mechanics - Projectiles Q's having problems

AI Thread Summary
To solve the projectile motion problem of a car driven off a cliff, the initial horizontal velocity is given as 15 m/s, and the wreckage lands 45 m away from the cliff. The time of flight can be calculated using the equation x = v0t, resulting in t = 3 seconds. The vertical height can then be determined using the equation for vertical displacement, x = 0.5gt^2, yielding a height of approximately 44.1 m. The initial approach using vertical motion equations was unnecessary since horizontal motion alone sufficed for the time calculation. Understanding the separation of horizontal and vertical components is crucial in projectile motion problems.
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Homework Statement


In a film stunt a car is driven over a cliff at 15ms^-1, the ground at the top of the cliff being level. The wreckage is 45m horizontally from the foot of the cliff.

What is the vertical height that the car has fallen?


Homework Equations


Im not sure but I am guessing s=ut+1/2at^2 should be used here.


The Attempt at a Solution


I've attempted to use the equation above, using v=u+at to get a value for t and using v=0, a=-9.8 and u=15. Now after finding a value for t and plugging that into the s=... eqn I am not getting the right answer.

Am i going about this the wrong way? Thanks a lot for your help.
 
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Use x = x0 + v0t + .5at^2
The car is not accelerating so all you need is x = v0t. Plug in your values...

45 = 15t
t = 3 seconds

Using x = .5gt^2
x = .5 * (9.8) * 3^2
x = 44.1 m
 

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