Projectile Motion Using Conservation of Energy

[JoeTheKid]

Homework Statement

Exam review question concerning 2 dimensional motion, the professor did not hand out an answer key and I am just looking for an answer check.

A car launches at an angle of 30 degrees above the horizontal, off a cliff (50m), moving at an initial speed of 10m/s. Find the maximum height achieved from the launch point (using only conservation of energy). What will be its speed when it lands?

Data
Cliff height 50m
ramp degree 30
initial velocity 10m/s

Homework Equations

Conservation of energy K1+U1=K2+U2
V(y) = V(yi) - gt
y=yi +V(yi)t - 1/2gt^2

The Attempt at a Solution

Im unsure if i did this correctly it has been a few months since we last touched upon projectile motion, but i began with Conservation of Energy, identifying that all the cars kinetic energy is at the launch point and its potential energy will be somewhere above the ramp. So...
1/2MV^2=MGH
rearranged for a height value
height=V^2/2g
using 10m/s sin 30 = 5 m/s for the speed the car was moving when it left the ramp.
plugged 5m/s back into h=5 m/s^2/2(9.8m/s^s) to get a height value of 1.27m

for max height because we traveled 1.27m above the 50m cliff we were already on i added the two values together to get a max height of 51.27m

And for speed because we are in a parabolic arc identified that at 51.27m all our vertical velocity is now 0 and the car is now free falling under the influence of gravity

So i used the equation y=yi +v(yi)t - 1/2gt^2
got rid of yi and v(yi) because their values are zero, then rearranged for time

t= √2y/g = √2(51.27m)/9.8 = 3.23s
To be find speed just took 51.27m/3.23s and got an answer of 15.87m/s, which doesn't seem right because if instead I take mgh=1/2mv^2 and found the Velocity using the new height of 51.27m i get a velocity of 31.74 which is double the velocity i got using one dimensional equations.

Any insight into this problem would be greatly appreciated!

 

[Simon Bridge]

... a height value of 1.27m
for max height because we traveled 1.27m above the 50m cliff we were already on i added the two values together to get a max height of 51.27m

You were asked for the height above the launch point - which is the first of those numers.

The way to think about conservation of energy is that the initial (vertical) kinetic energy goes entirely to changing the height.
So you have $\frac{1}{2}mv\sin\theta = mg\Delta h$ ... then your notation is cearer and, in your head, you are automatically thinking of the change in height rather than an absolute height.

So i used the equation y=yi +v(yi)t - 1/2gt^2

... you are instructed to use conservation of energy - so the use of this equation will lose you marks.

Just reverse the previous calculation - what distance does the car fall?