Physics Help Please?

Question

Benjamin fires an arrow from a castle wall 10m high. It leaves the bow with a speed of 40m/s directed 37° above the horizontal.

 

a) Find the initial velocity components

 

b) Find the maximum height of the arrow

 

c) Where does the arrow land? (Assume the ground is level beyond the base of the castle wall)

Answer 1

a) The formulas are

 

v_y0 = v₀ sin θ

 

v_x0 = v₀ cos θ

 

Plugging in v₀ = 40 m/s and θ = 37 deg, we have

 

v_y0 = 24.07 m/s

 

v_x0 = 31.95 m/s

 

b) When the arrow reaches its highest point, its vertical velocity (v_y) is zero. The formula is

 

v_y² = v_y0² + 2a_yΔy

 

The point where v_y = 0 corresponds to a specific Δy (the height the arrow goes relative to the archer on top of the castle); v_y0 was found in part (a), and a_y = -g.

 

Δy = v_y0² / 2g = (24.07 m/s)² / (2*9.8 m/s²) = 29.56 m

 

Measured relative to the ground, the arrow reaches a height of 10 m + 29.56 m = 39.56 m.

 

c) Since the final y position of the arrow is different from its initial position, the process for finding this is a bit more tedious, but still possible. I'll just quote the answer: https://en.wikipedia.org/wiki/Range_of_a_projectile. Using my notation, this is

 

Δx = (v_x0 / g)(v_y0 + sqrt[v_y0² + 2gy₀]

 

where y₀ = 10 m. Plugging in the values gives 169.26 m.