Question

A cannon shoots a cannonball from the ground level (ignore the height of the cannon) towards a cliff of height h = 170 m. The cannonball is launched with an initial velocity of 110 m/s at an angle of 64° above the horizontal. Neglect air resistance. Assume the cannonball hits the cliff as it descends (on its way down) exactly at the edge, as shown. 

a. Determine the maximum height above the ground reached by the cannonball.

b. What speed will the cannonball have at the maximum height? 

c. Determine the time it takes the cannonball to hit the cliff. 

d. How far away from the base of the cliff is the cannon? 

e. What will be the speed of the ball as it hits the cliff? 

f. What is the direction of the velocity of the ball as it hits the cliff?

Answer 1

a)

H = ( u sin x) ^2 / ( 2g) = (110* sin 64)^2 / 19.6 = 498.711 m

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b)

only horizontal component of velocity acts at highest point

Vx = 110 cos 64 = 48.22 m/s

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c)

using 2nd equation of motion

h = (u sin x) t - 0.5 g t^2

170 = ( 110 sin 64) t - 4.9 t^2

- 4.9 t^2 + 98.867 t - 170 = 0

solving for t

t = 18.279 s

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d)

x = vx * t = 881.42 m

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e)

v^2 = u^2 + 2 g h

110^2 = u^2 + 2 * 9.8* 170

u = 93.638 m/s

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f)

x = arccos ( 48.22 / 93.638) = 59

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