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A body is thrown vertically upwards with a speed 29.4m/s.At the same time another body...
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Let us say the two objects meets at x meters from ground at time t.
For the object that thrown upward;
`uarr S = ut+1/2at^2`
`x = 29.4t-1/2*9.81*t^2` -----(1)
For the object that dropped downward;
`darr S = ut+1/2at^2`
`44-x = 0+1/2*9.81*t^2` -----(2)
`44 = 29.4t`
`t = 1.497S`
`x = 33m`
So the two objects will meet at 33m height from the ground and they will meet after 1.497 seconds after they were thrown/dropped.
The upward throw starts at ground level
Posted by jeew-m on December 11, 2012 at 6:16 PM (Answer #1)
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