1. A cannon shell is fired at ground level such that the horizontal velocity is

Question

1. A cannon shell is fired at ground level such that the horizontal velocity is 40 m/sec and the vertical velocity up is 20 m/sec. If the air resistance is negligible: (1 Mark each) a) What is the maximum height of the cannon ball? b) What is the velocity of the cannon ball at this height? c) What is the time required to reach this height? d) What is the direction of the of the cannon ball's acceleration immediately upon firing? e) What is the horizontal distance travelled by the cannon ball? f) What is the cannon ball's final velocity?

Explanation / Answer

here,

1)

the horizontal velocity , vx = 40 m/s

vertical velocity , vy = 20 m/s

a)

the maximum height , hm = vy^2 /2g

hm = 20^2 /( 2 * 9.81) m

hm = 20.4 m

b)

at maximum height

the vertical velocity is zero

so, the velocity of shell is 40 m/s

c)

the time required to reach this height , t = vy /g

t = 20 /9.81= 2.04 s

d)

the direction of cannon ball accelration is dowanwards

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