An arrow is shot at an angle 20 degrees from the horizontal. If its height is increasing at 250 km/hr, how fast is the arrow itself going? (Not the horizontal speed, but it's actual diagonal speed.)

Answer 1

Hi Maddie H.,

A projectile with velocity v projected at an angle θ from the horizontal, has velocity in the x-direction v_x, and velocity in the y-direction v_y.

The velocity in the x-direction is v_x = v*cos(θ), and the velocity in the y-direction is v_y = v*sin(θ).

You are given the velocity in the y-direction of v_y = 250 km/hr, and angle θ to the horizontal of θ = 20^o.

We can set up the equation v_y = v*sin(θ) and solve for v:

250 km/hr = v*sin(20^o)

v = 250/sin(20^o) km/hr = 730.95 km/hr

I hope this helps, Joe.