An arrow is launched from P with a speed Vi = 25m / s. Knowing that the target Q is 10 m high, and the arrow reaches it as shown in the figure, we are asked to determine the distance X. (g = 10m / s2).

20 m

Explanation:

Given in the y direction:

Δy = 10 m

v₀ = 25 m/s sin 37° = 15.0 m/s

a = -10 m/s²

Find: t

Δy = v₀ t + ½ at²

10 m = (15.0 m/s) t + ½ (-10 m/s²) t²

10 = 15t − 5t²

2 = 3t − t²

t² − 3t + 2 = 0

(t − 1) (t − 2) = 0

t = 1 or 2

Since the projectile reaches Q before it reaches the peak, we want the lesser time, so t = 1.

Given in the x direction:

v₀ = 25 m/s cos 37° = 20.0 m/s

a = 0 m/s²

t = 1 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (20.0 m/s) (1 s) + ½ (0 m/s²) (1 s)²

Δx = 20 m