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Projectile Motion Example
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, va 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.  
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