An arrow is shot at an angle 37 ◦ with the horizontal. It has a velocity of 51 m/s. How high will the arrow go

What horizontal distance will it travel?
Answer in units of m

We need to consider the horizontal and the vertical component of velocity.
Vx and Vy.
Vx = V * cos(37) = 51 * cos(37) = 40.73 m/s

For the Vy component we need to include gravity too, which is -9.81 m/s² (because it acts downwards it is negative)

Vy= V*sin(37) - 9.81t

Now for vertical distance we integrate this formula:

Y= V*sin(37)*t - 4.9*t²
We need to know the maximum of Y so we differentiate this formula again (so you actually use the velocity equation)

dY/dt = V*sin(37) - 9.81*t = 0
Now solve for t => t = 3.13 s

Now substitute the time into the vertical distance equation:

Y(3.13) = 51 * sin (37) * 3.13 - 4.9*(3.13)² = 48.06 m

Now for the horizontal distance:

The arrow touches down when the vertical distance is 0 again, so we want the roots of the vertical distance equation:

So we want to solve Y = 0
Y = 51*sin(37)*t - 4.9*t² = 0
t (51*sin(37)-4.9t) = 0
t = 0 or 51*sin(37)-4.9t =0
t = 0 or t = 6.264 s

So Y is zero again at t=6.264 s
We know Vx=40.73 m/s

So in 6.264 seconds it travels X=40.73 * 6.264 = 255.12 m

Hope it helps!