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    Here is the link to the image http://img593.imageshack.us/img593/7647/questionb.jpg

    For part A, I already know that the answer is 16.5m. However, to find the initial velocity why do you have to do (3/5)*30=18ms-1 to get the initial velocity? Also would the normal velocity for this unknown?

    And for part B, to find distance I would have to use speed*time = distance.

    But in order to find the distance I would first have to find the time, So how would I have to find the time?

    Also when I find the time, I would have to multiply it to find the distance, So which speed would I have to use 30ms-1 or the initial velocity 18ms-1 and why?
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    The initial speed is made of a horizontal and a vertical component. The horizontal component is constant, as you are neglecting air resistance. However, the vertical component changes die to gravity, and therefore decelerates at 9.8 ms^-2.

    a) The initial vertical velocity = 30sinx. However, you are already told that sinx = 3/5, so the vertical speed is 30*0.6.

    b) Work out the time taken for the projectile to land using the vertical component ( just suvat)
    Once you have the time,use suvat horizontally (s=ut+ 0.5at^2, but as a=0, it's just s=ut). u is just the horizontal component of the speed, 30cos(arcsin0.6).
 
 
 
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Updated: September 28, 2011

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