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M2 - January Paper 2005 Projectiles Question. watch

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    A particle P is projected from a point A with speed 32m/s at an angle of elevation ? where sin = 3/5. The point 0 is on the horizontal ground, with 0 vertically below A and OA = 20m. The particle P moves freely under gravity and passes through the point B, which is 16m above the ground, before reach the ground at Point C.

    a) What is the time of the flight from A to C:4.8s
    b)the Distance OC: 120m
    C)the speed of P at B.
    Can someone guide me through part C please, explaining the logic behind it.
    (I have the mark scheme but it isn't helping.)
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    (Original post by k10k)
    A particle P is projected from a point A with speed 32m/s at an angle of elevation ? where sin = 3/5. The point 0 is on the horizontal ground, with 0 vertically below A and OA = 20m. The particle P moves freely under gravity and passes through the point B, which is 16m above the ground, before reach the ground at Point C.

    a) What is the time of the flight from A to C:4.8s
    b)the Distance OC: 120m
    C)the speed of P at B.
    Can someone guide me through part C please, explaining the logic behind it.
    (I have the mark scheme but it isn't helping.)

    well, cosθ=0.8 and tanθ=0.75..

    so veritcal speed = 0.6x32 = 19.2
    Horizontal speed = 0.8x32 = 25.6

    (c) Well, at B, the horizontal speed =25.6 as it doesn't slow down, and you can work out the vertical speed if you consider the vertical distance between A and B (i.e 4m) this works as the vertical speed at A is the same at any point where the particle is 20m above ground..

    v²=u²+2as

    v²=19.2²+(2x4x9.8)

    =>v=21.14 vertically

    then, for magnitude v => √25.6²+21.14² = 33.2m/s
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    oo, just checked the mark scheme posted on another thread and am correct, so if you need any more explanation i'll be happy to give
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    (Original post by k10k)
    A particle P is projected from a point A with speed 32m/s at an angle of elevation ? where sin = 3/5. The point 0 is on the horizontal ground, with 0 vertically below A and OA = 20m. The particle P moves freely under gravity and passes through the point B, which is 16m above the ground, before reach the ground at Point C.

    a) What is the time of the flight from A to C:4.8s
    b)the Distance OC: 120m
    C)the speed of P at B.
    Can someone guide me through part C please, explaining the logic behind it.
    (I have the mark scheme but it isn't helping.)
    if you resolve upwards you know that u=32sin?, a=-9.8 and s=-4. you can use constant acceleration formulae to work out v (ie v^2=u^2 +2as) you should get v=21.1 m/s

    since there is no horizontal acceleration, the horizontal velocity is 32cos?=25.6m/s

    now speed = magnitude of horizontal and vertical velocities = sqrt (21.1^2 + 25.5^2) = 33.2 m/s

    hope that helps
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    Are you going down? when using v^2=u^2+2as?
    (Kaizer)
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    yeah, think of the upsidedown U shape of the motion, and draw a horizontal line across from A until it meets the motion again...I use SUVAT from this point, as the vertical speed is the same...so the particle is accelerating downwards a distance of (20-16) metres..
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    Thank you Kaiser_Mole and Realicetic. Both of your methods make more sense now
    And good luck Friday, if you are taking M2 edexcel.
 
 
 
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