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    Hi guys

    Can anyone help me with this question? I can't seem to get the answer in the back of the book.

    It asks us to calculate the maximum height which I got correct - that was 23m approximately. It then asked us to calculate the horizontal displacement when the object lands on the ground. I got 45.88m.

    We were only given the information that the golf ball has a velocity of 30ms^'1 at an angle of 45 degrees to the horizontal ground.

    Any help would be greatly appreciated.
    Thanks
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    (Original post by Blake Jones)
    Hi guys

    Can anyone help me with this question? I can't seem to get the answer in the back of the book.

    It asks us to calculate the maximum height which I got correct - that was 23m approximately. It then asked us to calculate the horizontal displacement when the object lands on the ground. I got 45.88m.

    We were only given the information that the golf ball has a velocity of 30ms^'1 at an angle of 45 degrees to the horizontal ground.

    Any help would be greatly appreciated.
    Thanks
    Can you please post your working?
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    (Original post by Notnek)
    Can you please post your working?
    Sure, I got the velocity to be 21.21ms^-1 for both components.

    I then used v^2 = u^2 +2as where v=0 a= -9.81 and u = 21.21
    0 = 450 - 19.62s
    s = 450 / 19.62 which gave me 22.94m

    Then I used s=ut+0.5at^2 to get the time for the length of the whole projectile (so like how long it took to come back down to the ground)

    22.94 = 21.21t because a=0 at the maximum height
    therefore t=22.94/21.21 = 1.082 to get to max height
    I then doubled this for the time taken for the full projectile which gave me 2.1631

    Using s=ut (because a=0 horizontally) I then got s = 21.21 * 2.1631 = 45.88m
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    (Original post by Blake Jones)
    22.94 = 21.21t because a=0 at the maximum height
    This is incorrect. The acceleration due to gravity for the vertical motion is a constant : g = 9.8. It won't change throughout the vertical motion. The velocity does change and it is this that is equal to 0 at the apex.

    I recommend using s = ut + 1/2 at^2 with s = 0. This will give you the total time of flight (since when the particle hits the ground its vertical displacement will be 0).
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    (Original post by Notnek)
    This is incorrect. The acceleration due to gravity for the vertical motion is a constant : g = 9.8. It won't change throughout the vertical motion. The velocity does change and it is this that is equal to 0 at the apex.

    I recommend using s = ut + 1/2 at^2 with s = 0. This will give you the total time of flight (since when the particle hits the ground its vertical displacement will be 0).
    Oh! I see! Thank you!
 
 
 
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