A cannon on horizontal ground, at point C, is used to target a point T 25m behind a narrow wall. Unfortunately the cannon is damaged and can only fire at a 45 ∘ angle and at one speed. So, the only way to aim the cannon is by moving it towards and away from the target. The gunners aren't sure if they can actually hit the target. a. If the cannonball leaves the cannon at u =35ms −1 at what distance d must the cannon be placed in front of the wall in order to hit the target, if the wall is ignored and the target is at the same height as the cannon

b. The wall is 15.0m high. Does the cannonball actually go over the wall and hit the target? If so, by how much?

The projectile launch ratios allow to find the results on the cannonball launch questions are:

 a) The canyon is at x = 100 m from the wall.

 b) The bullet hits the wall at a height of 4.8 m above the height of the wall.

Projectile launching is an application of kinematics where there is no acceleration on the x-axis and the y-axis is gravity acceleration.

A) Indicate that the initial speed of the bullet is v = 35 m/s with an angle  θ=45º, which is the distance from the wall. The target is 25 m behind the wall, in the attachment we have a scheme of the system.

Let's find the range that is the horizontal distance for which the height is  zero.

Let's calculate.

       R =  

       R = 125 m

Therefore the canonl must be 100 m from the wall.

B) The height of the wall is y = 15m, the blah manages to pass it and by how much.

We use trigonometry to find the components of velocity.

         v₀ₓ = v₀ cos θ

          = v₀ sin θ

          v₀ₓ = 34 cos 45 = 24.7 m / s

          = vo sin 45 = 24.7 m / s

Let's find the time it takes for the cannonball to travel the x = 100 m

        t =  

        t = 4.04 s

Let's find the height for this time.

        y = t - ½ g t²

Let's calculate

        y = 24.7 4.04 - ½ 9.8 4.04²

        y = 19.8 m

This height is greater than the height of the wall h = 15m, therefore the cannoball passes the wall, by a difference in height.

        Δy = y -h

        Δy = 19.8-15

        Δy = 4.8 m

In conclusion, using the projectile launch ratios we can find the results on the cannonball launch questions are:

 a) The canyon is at x = 100 m from the wall.

 b) The cannonball hits the wall at a height of 4.8m above the height of the wall.

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Using the range formula   S = V^2 * sin 2 theta / g

So the cannonball should travel

S = 35^2 * sin 90 / 9.8 = 125 m

Then the cannon should be placed 100 m from the wall

H = Vy t - 1/2 g t^2   height of cannonball after time t

Vy = 35 * sin 45 = 24.7 m/s     vertical speed of cannonball

Vx = 35 * cos 45 = 24.7 m/s

t = 100 m / 24,7 m/s = 4.04 sec    time for cannonball to reach wall

H = Vy t - 1/2 g t^2       height of cannonball after time t

H = 24.7 * 4.04 - 1/2 * 9.8 * 4.04^2 = 19.8 m

19.8 m - 15 m = 4.8 m     cannonball clears wall by 4.8 m

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2