Projectile - Question 1

A ball is thrown upwards: The only motion is vertical. The ball is projected with a non-zero initial velocity and decelerates as the acceleration is -g. The maximum height is reached when the velocity = 0.
The ball is thrown vertically upwards at 15ms^-1 from a point 5 m above the ground. Calculate:(i) the maximum height above the ground(ii) the time taken to hit the ground(iii) the speed at which it hits the groundFor the maximum height above the ground use v² = u² + 2as to get s=11.5m.So the maximum height above the ground is 16.5m.To calculate the time taken to hit the ground uses= -5u=15v=..a= -9.81t=?Use s = ut + ½at² to give -5 = (15)t + ½(-9.81)t² giving

t =	15 + √[(15)2 - 4(4.905)(-5)]  2(4.905)

= 3.36sFor the speed at which it hits the ground use v² = u² + 2as = 15² + 2(-9.81)(-5) to get v = 18.0ms^-1.