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  # M2- Projectiles Help watch
  
  1. 17) A child throws a ball horizontally from a window 6m above horizontal ground. The ball just clears a vertical wall 2.5m high and 12m from the house. By modelling the ball as a particle calculate the speed of projection. The ball hits the ground at Q. Calculate the distance of Q from the wall.
  
  Vertically:
  s=12
  a=0
  
  Horizontally:
  s=2.5
  a=9.8
  u=0
  
  Resolving vertically: s=ut+1/2at^2
  t=5/7s
  
  Resolving horizontally: s=ut+1/2at^2
  12=5/7u
  u=16.8m/s
  
  b) Horizontally:
  v=16.8
  u=16.8
  a=0
  
  Vertically:u=0
  a=9.8
  s=5
  
  Resolving vertically: s=ut+1/2at^2
  5=4.9t^2
  t=1.01s
  
  Resolving horizontally: s=ut+1/2at^2
  =16.8*1.01
  =16.97
  
  Ergo, Horizontal distance from wall=16.97m-12=4.97
  
  The textbook has the answer of 4.97m. However, what I am asking is that it asks for the distance of Q from wall, this is not the horizontal disance. So shouldn't it be:
  
  Root of 4.97^2+2.5^2
  
  18) A tennis ball is served horizontally at a speed of 24m/s from a height of 2.7m. The net is 1m high and 12m horizontally from the server. Model the ball as a particle and hence determine whether the ball cclears the net and if so by what distance.
  
  How do you do 18? I have little idea.
  2. First - Didn't read all the working, but it just means the horizontal distance. That's the shortest distance to the wall. You worked out the distance to the top of it I think?)
  
  Second - So you can look at horizontal and vertical motion. You know horizontal distance and speed - can you see how to use this? And then how you can use what you find in the vertical motion?
  Spoiler:
  Show
  find the time at which the ball has gone 12m horizontally
  Spoiler:
  Show
  use this time and the acceleration of gravity to find how far it's fallen
  3. (Original post by gcseeeman)
  First - Didn't read all the working, but it just means the horizontal distance. That's the shortest distance to the wall. You worked out the distance to the top of it I think?)
  
  Second - So you can look at horizontal and vertical motion. You know horizontal distance and speed - can you see how to use this? And then how you can use what you find in the vertical motion?
  Spoiler:
  Show
  find the time at which the ball has gone 12m horizontally
  Spoiler:
  Show
  use this time and the acceleration of gravity to find how far it's fallen
  I did it as such:
  
  Horizontal:
  u=24
  v=24
  a=0
  s=12
  t=?
  
  s=ut+1/2at^2
  12=24t
  t=2
  
  Now what?
  4. (Original post by Deep456)
  I did it as such:
  
  Horizontal:
  u=24
  v=24
  a=0
  s=12
  t=?
  
  s=ut+1/2at^2
  12=24t
  t=2
  
  Now what?
  If 12=24t, t isn't 2 (I guess you see the silly mistake). But use that for vertical motion and see how far it falls in that time.
  5. (Original post by gcseeeman)
  If 12=24t, t isn't 2 (I guess you see the silly mistake). But use that for vertical motion and see how far it falls in that time.
  Ah yes, that was a school boy error.
  
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