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    How to solve this question?'A boy projects a ball vertically upwards with speed 10m/s from a point, X, which is 50m above the ground. T seconds after the ball is projected upwards, the boy drops a second ball from X. Initially the second ball is at rest. The balls collide 25m above the ground. Find the value of T.'I am not getting the motion of the balls. I've used SUVAT and got that as the first ball is thrown up in the air it must come down and therefore reach a peak where it's velocity is 0m/s which takes 1.02 seconds as u=10, s=5.1, v=0, a=-9.8 so t=1.02 so the time it takes to get back to being 50m above ground is 2.04 seconds.
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    (Original post by adriannala)
    How to solve this question?'A boy projects a ball vertically upwards with speed 10m/s from a point, X, which is 50m above the ground. T seconds after the ball is projected upwards, the boy drops a second ball from X. Initially the second ball is at rest. The balls collide 25m above the ground. Find the value of T.'I am not getting the motion of the balls. I've used SUVAT and got that as the first ball is thrown up in the air it must come down and therefore reach a peak where it's velocity is 0m/s which takes 1.02 seconds as u=10, s=5.1, v=0, a=-9.8 so t=1.02 so the time it takes to get back to being 50m above ground is 2.04 seconds.
    Calculate how long it would take the first ball to reach a point 25m above the ground (i.e. 25m below its initial start point).
    Calculate how long it would take the second ball to free fall 25m under gravity.
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    Hey, ok.

    So the first ball is thrown up with u 10m/s and will reach its peak when v=0m/s as the ball will stop moving. Gravity acts so a=-9.8m/s^2. So we have u, v and a we need t so can use v=u+at so 0=10+(-9.8)(t) so 9.8t=10 so t=1.02 so to get back down to 50m above the ground thats another 1.02 seconds making it 2.04 seconds. To get to -25m below (25m above the ground) it will be s=-25, a=9.8. We need u so we have to find this from the previous information and in the previous part s=5.1, u=0 (previous u, which will now be v), a =9.8 so we can use v^2=0+2x9.8x5.1 so v=9.99 basically 10. So the u is 10 just like for the other ball. Then you get a quadratic ten solve and ok. I get it thank you.
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    (Original post by adriannala)
    Hey, ok.

    So the first ball is thrown up with u 10m/s and will reach its peak when v=0m/s as the ball will stop moving. Gravity acts so a=-9.8m/s^2. So we have u, v and a we need t so can use v=u+at so 0=10+(-9.8)(t) so 9.8t=10 so t=1.02 so to get back down to 50m above the ground thats another 1.02 seconds making it 2.04 seconds. To get to -25m below (25m above the ground) it will be s=-25, a=9.8. We need u so we have to find this from the previous information and in the previous part s=5.1, u=0 (previous u, which will now be v), a =9.8 so we can use v^2=0+2x9.8x5.1 so v=9.99 basically 10. So the u is 10 just like for the other ball. Then you get a quadratic ten solve and ok. I get it thank you.
    Far more simply you can use the displacement formula
    s=ut+0.5at^2
    Choosing the positive direction to be down..........
    For the first ball s=25, u=-10 and a=g=9.8 and solve the resulting quadratic for t

    For the second ball, s=25, u=0 and a=9.8
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    (Original post by gdunne42)
    Far more simply you can use the displacement formula
    s=ut+0.5at^2
    Choosing the positive direction to be down..........
    For the first ball s=25, u=-10 and a=g=9.8 and solve the resulting quadratic for t

    For the second ball, s=25, u=0 and a=9.8
    Ok thank you!
 
 
 
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