Film stunt physics car problem

[kingyof2thejring]

In a film stunt, a car is driven off a cliff with a horizontal velocity of 12.9 ms-1. The cliff face is vertical, and the cliff is 20.7 m high. Calculate the distance in m from the base of the cliff to the point where the car strikes the ground.

should i be considering this problem in x and y - components of the motion separately.

taking upwards as +ve direction for vectors
displacment = -20.7
initial displacement = 0
inital speed = 12.9
a=-9.8

using it in s = ut + 1/2at^2 i get
20.7 + 12.9t - 4.9t^2
t = 3.757 +ve value taken

s=ut
12.9*3.757= 48.5m as the displacement
and then 48.5 + 20.7 = 69.2m for the distance.
help would be apprecated
thanks in advance

 

[BobG]

Only one mistake.

The horizontal velocity is 12.9 m/sec. There is no initial vertical velocity.

But, your overall approach of solving the x component and y component separately is right. Solve the y component to find the time. Use that time to solve the horizontal component.

 

[kingyof2thejring]

cheers!