Question

2. (a) A ball is fired from a cannon straight up at a velocity of 30ms^ - 1. The ball leaves the cannon at a height of 5 m that is vertically above the ground. Find the time taken until the ball hits the ground. Find the velocity of the ball when it hits the ground. If the ball had been projected at 10 ms^ - 1 instead of 30 ms^ - 1, what would its final velocity when it hits the ground have been? (Use g = 10 ms^ -1.)

Answer 1

Here ,

velocity , v = 30 m/s

hi = 5 m

let the time taken is t ,

using second equation of motion

-5 = 30 * t - 0.5 * 9.8 * t^2

solving for t

t = 6.28 s

the time taken by ball to hit the ground is 6.28 s

let the velocity at the ground is v

v = 30 - 6.28 * 9.8

v = -31.54 m/s

the velocity of the ball is 31.54 m/s downwards

----------------------------------

for ui = 10 m/s

using third equation of motion

v^2 - 10^2 = 2 * 9.8 * 5

v = 14.1 m/s

the final speed of the ball would be 14.1 m/s