assume that the initial speed is 25 m/s and the angle of projection is 53 degree above the hroizontal. the cannon ball leaves the uzzel of the cannon at a highet of 200 m.( the cannon is at the edge of the cliff) A: find the horizontal distance the cannon travles. B: when does the cannon ball reach the ground? C: find the maximum highet the cannon ball reaches.

A.  xmax = 131.49 m

B.  t = 8.74 s

C.  ymax = 220.33 m

Explanation:

A. In order to find the horizontal distance which cannon travels you first calculate the flight time. The flight time can be calculated by using the following formula:

     (1)

yo: height from the projectile is fired = 200m

vo: initial velocity of the projectile = 25m/s

g: gravitational acceleration = 9.8 m/s^2

θ: angle between the direction of the initial motion of the ball and the horizontal = 53°

t: time

You need the value of t when the projectile hits the ground. Then, in th equation (1) you make y = 0m.

When you replace the values of all parameters in the equation (1), you obtain the following quadratic formula:

(2)

You use the quadratic formula to obtain the value of t:

You use the positive value because it has physical meaning.

Now, you can calculate the horizontal range of the projectile by using the following formula:

The cannon ball travels a horizontal distance of 131.49 m

B. The cannon ball reaches the canon for t = 8.74s

C. The maximum height is obtained by using the following formula:

    (3)

By replacing in the equation (3) the values of all parameters you obtain:

The maximum height reached by the cannon ball is 220.33m

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

c. newton's third law of motion