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    The Q is Karen is standing 4m, away from a wall which is 2.5M high, she thrpows that ball at 10ms at an angle of 40 degrees, to the horizontal from a heigh of 1m above the ground, will the ball pass over the wall? What am i suppose to find, I thought it would be the max heigh minus 2.5m but thats not right
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    (Original post by SunDun111)
    The Q is Karen is standing 4m, away from a wall which is 2.5M high, she thrpows that ball at 10ms at an angle of 40 degrees, to the horizontal from a heigh of 1m above the ground, will the ball pass over the wall? What am i suppose to find, I thought it would be the max heigh minus 2.5m but thats not right
    Once the ball has travelled a horizontal distance of 4 metres, what is the total height of the ball? Is it bigger than 2.5? If so, then it does pass over. If not, then no.
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    (Original post by SunDun111)
    The Q is Karen is standing 4m, away from a wall which is 2.5M high, she thrpows that ball at 10ms at an angle of 40 degrees, to the horizontal from a heigh of 1m above the ground, will the ball pass over the wall? What am i suppose to find, I thought it would be the max heigh minus 2.5m but thats not right
    To elaborate on my previous reply, you'll need to do this:

    (i) set up an equation for horizontal motion, this will be x = 10t cos 40^{\circ}.

    (ii) find the time at which the ball has moved 4 metres horizontally by solving 4 = 10t \cos 40^{\circ}}

    (iii) set up an equation for vertical motion, more specifically, one to determine vertical height; this will be s = ut \sin \theta - \frac{1}{g}t^2

    (iv) you know t from (ii) - plug this into the equation for (iii) and see what the vertical displacement is.

    (v) add 1 to your answer from (iv) because the ball was launched off the ground

    (whatever comes after v) if the answer to (v) is bigger than 2.5, it passes over the wall; if not, then it doesn't.
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    (Original post by Zacken)
    To elaborate on my previous reply, you'll need to do this:

    (i) set up an equation for horizontal motion, this will be x = 10t cos 40^{\circ}.

    (ii) find the time at which the ball has moved 4 metres horizontally by solving 4 = 10t \cos 40^{\circ}}

    (iii) set up an equation for vertical motion, more specifically, one to determine vertical height; this will be s = ut \sin \theta - \frac{1}{g}t^2

    (iv) you know t from (ii) - plug this into the equation for (iii) and see what the vertical displacement is.

    (v) add 1 to your answer from (iv) because the ball was launched off the ground

    (whatever comes after v) if the answer to (v) is bigger than 2.5, it passes over the wall; if not, then it doesn't.
    I worked it out before you answered, It was pretty simple, dont know why i struggled, also I finally understood the question that I asked you yesterday thank you,

    Quick question, If a stone is thrown off a cliff downwards how do you work out how high the cliiff is?
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    (Original post by SunDun111)
    I worked it out before you answered, It was pretty simple, dont know why i struggled, also I finally understood the question that I asked you yesterday thank you,

    Quick question, If a stone is thrown off a cliff downwards how do you work out how high the cliiff is?
    Well, you'll need more information - but if you know the time and initial velocity, you can use h = ut - \frac{1}{2}gt^2.

    What question did you ask me yesterday? :-)
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    (Original post by Zacken)
    Well, you'll need more information - but if you know the time and initial velocity, you can use h = ut - \frac{1}{2}gt^2.

    What question did you ask me yesterday? :-)
    the one about changing a number into a surd

    Oh the info is, the ball is thrown with speed 10ms at an angle of projecttion 30 degrees from the cliff, hitting the sea at 2.5 seconds
    how high would the cliff be?
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    (Original post by SunDun111)
    the one about changing a number into a surd

    Oh the info is, the ball is thrown with speed 10ms at an angle of projecttion 30 degrees from the cliff, hitting the sea at 2.5 seconds
    how high would the cliff be?
    Just use the equation I put above - you know the time, the initial speed, the angle of projection - you can work vertically and find the vertical displacement using s = ut\sin \theta - \frac{1}{2}gt^2
 
 
 
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