1. ## Parametrics and Vectors

If a ball is thrown at a velocity of 75 ft/sec at an angle of 25 degrees with horizontal, and the ball is released 5 feet above ground, how do I find the angle needed when X = 215?

I Set the equation 215 = (75 cos X)t but there are two variables so I don't know how to solve it.

If an airplane flying at an altitiude of 3500 feet is dropping supplies, the path of the plane is parallel to the ground as the supplies are released while it travels 300mph, what are the parametric equations that represents this?

2. Originally Posted by Alan306090
If a ball is thrown at a velocity of 75 ft/sec at an angle of 25 degrees with horizontal, and the ball is released 5 feet above ground, how do I find the angle needed when X = 215?
This makes no sense at all if you don't tell us what "X= 215" means! If you mean that X is the distance to the point where the ball hits the ground, tell us that!

I Set the equation 215 = (75 cos X)t but there are two variables so I don't know how to solve it.
What angle? You say in the problem that the ball is thrown at an angle of 25 degrees so presumably that is not the angle you are talking about. Go back and read the problem carefully.

If an airplane flying at an altitiude of 3500 feet is dropping supplies, the path of the plane is parallel to the ground as the supplies are released while it travels 300mph, what are the parametric equations that represents this?
Pretty much what you have with 3500 feet added to the vertical. The initial motion of the supplies is the same as the initial motion of the airplane so it is 300 mph horizontally, 0 vertically.

x= (300 mph) t, $y= -(16)t^2+ 3500$

You will need to think about your units of measurement. If t is in seconds or minutes and you are measuring distance in feet, you will have to convert "300 mph" to "feet per minute" or "feet per second" before doing any calculation.