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, 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.

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.

To find the range of the projectile, you need to calculate the time it takes for the projectile to reach the ground. Since the initial velocity only has vertical and horizontal components and there is no horizontal acceleration, the time of flight will be the same for both vertical and horizontal motion.

To find the time of flight, you can use the y-component of the initial velocity and the acceleration due to gravity. The acceleration due to gravity is typically represented as "g" and is approximately 9.8 m/s^2. However, since the projectile is moving upwards initially, you need to take the negative value of the acceleration due to gravity:

t = V[sub]Y[/sub] / (-g)
where V[sub]Y[/sub] is the vertical component of the initial velocity, which you calculated to be 39.11 m/s.

Now, you can find the time of flight:

t = 39.11 m/s / (-9.8 m/s^2)
t = -3.9898 s
(Note: The negative sign indicates that the time is in the upward direction.)

Since the time of flight is negative, it suggests there is a problem with your calculations. After checking your formulas, I realized that you didn't account for the negative sign of the y-component. The correct value should be:

V[sub]Y[/sub] = -39.11 m/s (negative because the motion is upwards)

Now, you can recalculate the time of flight:

t = -39.11 m/s / (-9.8 m/s^2)
t = 3.9898 s

Now that you have the time of flight, you can find the range (horizontal distance traveled) by multiplying the horizontal component of the initial velocity (V[sub]X[/sub]) by the time of flight:

Range = V[sub]X[/sub] * t
where V[sub]X[/sub] is the horizontal component of the initial velocity, which you calculated to be 51.91 m/s.

Range = 51.91 m/s * 3.9898 s
Range = 207.05 meters

So, the range of the projectile is approximately 207.05 meters.