An arrow is shot from a bow at 20.0 m/s at an angle of 65 degrees above the horizontal. The arrow leaves the bow at a height of 1.80 m. At what height will it strike a wall that is 10.0 m away? (Remember to connect the two motions of the arrow using time.)

Answer 1

y = h + tanθ x - g/(2(vcosθ)^2) x^2
so, plug in x=10:

y = 1.8 + 2.144x - .580x^2
= 1.8+21.44-5.80
= 17.44

you can also use time, but it takes a bit more work:

Vx = 20 cosθ
use that to find t = 10/Vx
Vy = 20 sinθ
y = 1.8 + Vy t - 4.9t^2

If you churn it out, it should come out the same.