Basic Mechanics - Projectiles Q's having problems

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SUMMARY

The discussion focuses on solving a projectile motion problem involving a car driven off a cliff at an initial velocity of 15 m/s. The horizontal distance traveled by the car is 45 meters, leading to the calculation of the vertical height fallen. The correct approach involves using the equations of motion: first, determining the time of flight with the equation x = v0t, yielding t = 3 seconds, and then applying the vertical motion equation x = 0.5gt² to find the height, resulting in approximately 44.1 meters.

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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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