A cannonball is fired from a cannon at 50.0 m/s, at an angle of 30.0° above the horizontal on level ground.

Question

A cannonball is fired from a cannon at 50.0 m/s, at an angle of 30.0° above the horizontal on level ground. The goal is to determine the distance from its starting point to its ending point. The path of the cannonball is shown below. (inverse parabola)

 

(a) Determine how long it takes for the cannonball to hit the ground. 

(b) Calculate Δx. 

Answer 1

Hi Mia,

 

(a) For the first part

 

   Y = V_0yt -0.5gt²   since the cannonball hits  the ground Y=0

 

  0 = 50*sin30*t - 0.5*9.8*t²     because V_0y = 50sin30

  0 = 25*t -4.9*t²

  0 = (25 - 4.9t)t

 

t = 0 before firing the cannon  and t =25/4.9 = 5.1 seconds when the cannonball hit the ground

 

(b) ΔX is the difference between the range of the cannonball and the calculated horizontal distance

 

      R = V₀² sin2(θ)/g = 50²*sin60/9.8 =220.92 m

 

      X = V_0x * t = 50cos30*5.1 = 220.84 m

 

     ΔX = R - X =220.92 - 220.84 = 0.10 m

 

     ΔX = 10 cm

    

Answer 2

The velocity, 50 m/s, has two components - vertical and horizontal velocities.

The vertical component = 50 sin 30 = 25 m/s

The horizontal component = 50 cos 30 = 43.3 m/s

(a)  Let t be the time taken for the vertical component to reach its peak from initial velocity = 25 m/s to its final velocity = 0.

      Using the linear motion equation v = u - gt

      0 = 25 - 10t

      t = 2.5 s

     Time taken to go up and down = 5 s

     Time to hit the ground = 5 s

(b) Horizontal distance delt x = 43.3 * 5 = 216.5 m