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## Homework Statement

Exam review question concerning 2 dimensional motion, the professor did not hand out an answer key and Im 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!