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?

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

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.

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2