A small metal ball is thrown upwards, at 30° to the horizontal, with an initial
speed of 20ms^{1} 

Assuming that air resistance is negligible, calculate 

i) the time taken
by the ball to reach its highest point 

ii) the maximum height reached 

iii) the horizontal distance moved by the ball when it hits the ground
(assuming it is thrown above a horizontal surface) 

iv) the magnitude and
direction of the velocity of the ball 1.6s after it was thrown. 



i) At maximum height, the vertical component of the
velocity is zero; the ball is, of course, still moving but only horizontally. 



To find the time taken we use 



with v = 0 and u = 20cos60 



Time to reach maximum height is t = 1.02s 



ii) The average value, v_{a} of the vertical component of the
velocity of the ball is 



therefore the maximum height reached is 



Maximum height reached is h = 5.1m 



iii) Remembering that the total time in the air is 2t and that the horizontal
speed is constant we have 



so, the range of the projectile on a horizontal plane is x = 35.3m 



iv) We know that the magnitude of the horizontal component does
not change. So if we find the magnitude of the vertical
component of the velocity at t = 1.6s, a simple vector addition will give us
the answer. 

Considering the vertical component only: 



so, at t = 1.6s, the vertical component of velocity is v = 5.68ms^{1} downwards. 

The horizontal component of the velocity is
20cos30 = 17.32ms^{1} 



Inspection of the vector diagram here gives us 



and 



Therefore, at t = 1.6s, the ball is moving at 18.23ms^{1}
at 18.2° to the horizontal. 
