How would I solve the following problems?

Question

An airplane, diving at an angle of 48.0° with the vertical, releases a projectile at an altitude of 760 m. The projectile hits the ground 6.00 s after being released. (Assume a coordinate system in which the airplane is moving in the positive horizontal direction, and in the negative vertical direction. Neglect air resistance.)
(a) What is the speed of the aircraft in m/s?
(b) How far did the projectile travel horizontally during its flight in meters?
(c) What were the horizontal and vertical components of its velocity just before striking the ground in m/s?

I realize this is a school homework question and I've been trying to solve it since last night with the formulas of projectile motion and I cannot get the correct answer and don't know why. So could someone please help me out.

Answer 1

This sounds suspiciously like a school homework assignment.It is not a difficult question.I think you would be better served if you attempted to do it yourself. Good Luck!

Answer 2

Use the equations of motion

for the projectile, the initial v is the speed of the aircraft and
y(t)=760-v*cos(48)*t-.5*g*t^2

you know that y(6)=0
solve for v
0=760-v*cos(48)*6-.5*9.8*36
v=145 m/s

the horizontal distance is
x(6)=v*sin(48)*6
=647 meters

the components of v at impact are
vx=145*sin(48)
vy=145*cos(48)+9.8*6


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