A ball is thrown vertically with a an initial velocity of 30ms^-1. One second later, another ball is thrown upwards with an initial velocity of u ms^-1. The balls collide after a further 2 seconds. Find the value of u.
please explain in steps, thank you
please explain in steps, thank you
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For the first ball:
s = ut + 0.5at^2
s = (30 x 3) + (0.5 x -9.81 x 3^2)
s = 45.855m = height of first ball after 3 seconds
For the second ball:
s = ut + 0.5at^2
45.855 = 2u + (0.5 x -9.81 x 2^2)
2u = 65.475
u = 32.74m/s
s = ut + 0.5at^2
s = (30 x 3) + (0.5 x -9.81 x 3^2)
s = 45.855m = height of first ball after 3 seconds
For the second ball:
s = ut + 0.5at^2
45.855 = 2u + (0.5 x -9.81 x 2^2)
2u = 65.475
u = 32.74m/s
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deltaX1 = (1/2)*g*t^2 + Vi*t ===> here is : t=3 sec and Vi = 30 & g=9.8
deltaX2 = (1/2)*g*t^2 + Vi*t ===> here is : t=2 sec and Vi =U & g=9.8
because the balls collide, then deltax1 = deltax2
and replace t and Vi & g , in above deltaX s ,and you can find U
:)
deltaX2 = (1/2)*g*t^2 + Vi*t ===> here is : t=2 sec and Vi =U & g=9.8
because the balls collide, then deltax1 = deltax2
and replace t and Vi & g , in above deltaX s ,and you can find U
:)