Question
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
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
Answers
Steve
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.
find t when y=0.
Then multiply that by Vx. That's how far it went horizontally while going up and back down.
Hanky
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.