Recent content by RjD12

Homework Statement Mortar crew is near the top of a steep hill. They have a mortar. They angle this mortar at an angle of \theta = 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 \phi =...

Okay. Yes, that looks like what I have after fixing it. I must be creating some sort of error in my calculations then, because I plugged the equation : -4.9t2 + 4sin(42)*t +3 into a graphing calculator, and there is indeed a positive value. It is odd. Using the quadratic formula, underneath...

So what would be the final Y position? It seems like since it is in motion, it would start at zero.

If I use y= y0 + V0yt - (0.5)gt2 I get 12= 0 + 4sin(42)t-(9.8/2)t2 In the general equation form, it is: -4.9t2+4sin(42)-12 Unless this is incorrect, my answer is negative under the radical.

I also recently tried using the quadratic formula with my Y equation, but I end up getting a negative under the radical. I am not sure why.

Would I have to set them equal to each other by solving for time? I know that the time is the same for both directions.

They both have initial velocities, and those can be figured out using the velocity given and the angle. X0 would be 4cos(42) and Y0 would be 4sin(42). Am I getting any closer?

Homework Statement At a time t, a projectile is measured to have a height Y = 12 m above the ground and a velocity v = 4 m/s at an angle (theta) = 42 with respect to to the horizontal. At what horizontal positions is the particle at a height 3/4 Y ? Height= 12 m Velocity= 4 m/s Angle= 42...