# physics

posted by
**nix** on
.

an airplane is diving at a speed of 300 km/h and the angle betweent he plane's dive path and the vertical is 50 degrees. The plane launches a rocket which lands at 750m horizontally from beneath the launch point. a) find the flight time of the rocket and b) how high was the release point?

I would need to convert the 300 km/h into m/s to match the 750m but i don't know the equations i should use.

Take the horizontal velocity of the plane as the horizontal velocity of the rocket (odd rocket))(VcosTheta).

Take the initial vertical veloicty of the rocket as the same as the vertical velocity of the plane (VsinTheta)

vertical

h= VSinTheta*time +1/2 g time^2

horizontal

distance=VcosTheta*time

solve for time in the second equation, put it in the first equation.