Question

Projectile motion challenge problem. You are at war. You are stationed with a cannon that can only fire shells at 100 m/s. You may adjust the angle but it takes five seconds to do so. You are positioned on a strategically important bridge 70m high with a mission to protect it from the enemy. Spotters have alerted you to a remote controlled speedboat 3000 m away heading West to your position. It is carrying a bomb to blow up the bridge. The spotter tells you that its initial velocity is 26 m/s and its accelerating at 11 m/s2. You must fire the cannon and destroy the boat as fast as possible before it reaches the bridge. When the speedboat is 3000 m away set that as time = 0s. A. What time do you fire and what is the angle when you do fire? B. Right before you fire a 10m/s wind starts blowing from behind you towards the East. How do you adjust your fire?

Answer 1

A. the time of fire is 15 seconds, and the angle of fire is approximately 23.794° above the horizontal

B. The angle of fire is increased to approximately 35.4126° above the horizontal

Step-by-step explanation:

A. The height of the bridge, h = 70 m

The speed with of the shell, v₀ = 100 m/s

The location of the speedboat = 3000 m

The direction of the speedboat = 26 m/s

The acceleration of the speedboat, a = 11 m/s²

Let t represent the time of firing the shells, and let x represent the distance of the speedboat from the bridge, and let θ, represent the angle to fire with, we have;

For the speedboat, t = x/(100 × cos(θ))

We note that the time the shell can travel the 3,000 m = 30 seconds

Therefore an adequate time to fire is, t = 15 seconds

The distance the speedboat covers in 15 seconds is given as follows;

s = u·t + 1/2·a·t²

s = 26 × 15 + 1/2 × 11 × 15² = 1627.5 m

At the point the speedboat had traveled 1,627.5 m, the distance of the speedboat, x is then 3000 - 1,627.5 = 1,372.5 m from the bridge, the angle of fire is given from the following formula;

t = x/(100 × cos(θ))

15 = 1,372.5/(100 × cos(θ))

cos(θ) = 1,372.5/(15 × 100) = 0.915

θ = cos⁻¹(0.915) ≈ 23.794°

The angle of fire, θ ≈ 23.794° above the horizontal in the direction of the speedboat

B. Given that a 10 m/s wind is blowing towards East, we have;

The horizontal velocity towards East = 10 + v₀ × cos(θ)

The angle of firing is therefore, given as follows;

15 = 1,3725.5/(10 + 100 × cos(θ))

(10 + 100 × cos(θ)) = 1,3725.5/15

100 × cos(θ) = 1,372.5/15 - 10

cos(θ) = (1,372.5/15 - 10)/100 = 0.815

θ = cos⁻¹(0.815) ≈ 35.4126°

Therefore, the angle of fire, θ, will be increased to approximately 35.4126° above the horizontal in the remote controlled speedboat direction.