well, y(t) = 39.11t - 4.9t^2
find t when y=0.
Then multiply that by Vx. That's how far it went horizontally while going up and back down.
A projectile is fired at 37.0° above the horizontal with an initial speed of 65.0 m/s. Find it's range.
Well what I did was get the x and y components for the vector by:
V-subX= (65 m/s) (cos(37°)) = 51.91
V-subY= (65 m/s) (sin(37°)) = 39.11
I'm not sure where to go from here haha
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Steve, thank you for replying. I'm sorry but I'm not exactly sure why you used the function y(t). also I don't know if that's a specialized equation. We are only supposed to use the 4 kinematic equations.
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