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A particle projected horizontally with a speed `sqrt2gT` , from a height h above the...
A particle projected horizontally with a speed `sqrt2gT` , from a height h above the ground,where T is a constant, moves under gravity.
If the particle is at a distance (3/2) gT^2 from the point of projection, When it falls to the ground, show that, using the speed-time graphs, the time taken by the particle to reach the ground is T, and h=(1/2)gT^2.
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It is given that the particle is at a distance `(3/2) gT^2` from the point of projection, When it falls to the ground.
Area of the horizontal velocity time graph `= sqrt2gTxxt`
Horizontal distance travelled by the graph `= sqrt2gTxxt`
For the vertical velocity time graph;
`h = (0+g(t))/2xxt`
`h = g(t^2)/2`
Now we got that `h = g(t^2)/2` ,horizontal distance travelled by the particle is `sqrt2gTxxt` and when the particle hit the ground it is `3/2gT^2` from the point of projection. These three forms a pythogorous triangle.
`(g(t^2)/2)^2+(sqrt2gTxxt)^2 = (3/2gT^2)^2`
`t^4/4+2T^2t^2 = 9/4T^4`
`t^4+8T^2t^2-9T^4 = 0`
`t^4+9T^2t^2-T^2t^2-9T^4 = 0`
`t^2(t^2+9T^2)-T^2(t^2+9T^2) = 0`
`(t^2-T^2)(t^2+9T^2) = 0`
`(t-T)(t+T)(t^2+9T^2) = 0`
Since T is constant and t is positive;
`t = T`
`h = gT^2/2`
So the answers are proved.
Posted by jeew-m on August 8, 2013 at 11:30 PM (Answer #1)
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