A cannon as shown in the figure below is fired from Xo distance to the cliff wit

Question

A cannon as shown in the figure below is fired from Xo distance to the cliff with height h. Initial velocity of the cannon is 500 m/s and it leaves the cannon with an angle of 30 degrees. After 52.9 seconds cannon ball hit the ground at 15.13km from the cliff edge. (ignore the height of the cannon and air resistance)

b) At what angle the cannon ball should be fired to hit a target at 25 km far.

please show me all the steps

1) A cannon as shown in the figure below is fired from xo distance to the cliff with height h. Initial velocity of the cannon ball is 500 m/s and it leaves the cannon with an angle of 30o. After 52.9 seconds cannon ball hit the ground at 15.13 km from the cliff edge. (ignore the height of the cannon and air resistance) a) Determine the distance from the cliff (xo) and the height, h, of the cliff b) At what angle the cannon ball should be fired to hit a target at 25 km far. 500 m/s 30o Xo 5.13 km

Explanation / Answer

here,

initial speed , u = 500 m/s

time of flight , t = 52.9 s

theta = 30 degree

a)

(x0 + 15130) = u * cos(theta) * t

(x0 + 15130) = 500 * cos(30) * 52.9

x0 = 7776 m

the height , h = - u * sin(theta) * t + 0.5 * g * t^2

h = - 500 * sin(30) * 52.9 + 0.5 * 9.8 * 52.9^2

h = 487.2 m

b)

for x = 25000 m

let the angle be phi

x = u * cos(phi) * t

25000 = 500 * cos(phi) * 52.9

phi = 19.1 degree

the angle must be 19.1 degree

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