Mortar crew is near the top of a steep hill. They have a mortar. They angle this mortar at an angle of [itex]\theta[/itex] = 65° . The crew fires a shell at a muzzle velocity of 228 ft/sec (69.5 m/s). How far down the hill does the shell strike if the hill subtends an angle of [itex]\phi[/itex] = 32° from the horizontal?
How long will the mortar remain in the air?
How fast will the shell be traveling when it hits the ground?
Relevant diagram: http://ift.tt/1m5d84y
2. Relevant equations
X= Xo + Voxt
Y= Yo + Voyt – (1/2)at2
3. The attempt at a solution
First off, I’m not expecting to get all of my questions answered. I just need a little push.
I’m not sure where to start off at here. The fact that there are two angles here confuses me in regards to how they work in the equations.
I can say that Vox = 69.5cos(65) and that Voy = 69.5sin(65).
I’m really thrown off by the way the angles work here, and whether the distance works with a simple range equation. Any tips on where to start?
edit: Additionally, I was given the equation d = Vo + (1/2)at2 as a hint for this. Isn’t this wrong though, seeing as how the velocity should be multiplied with time?