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Two balls are thrown up simultaneously from the top of a 400 m high tower with speed 10 m//s and 55 m//s. Sketch a graph showing the separation between the balls versus time. |
Answer» SOLUTION :Taking O as the origin, displacement of ball(1), `s_(1)=400+10t-1/2 g t^(2)` Displacement of ball (2), `s_(2)=400+55t-1/2 g t^(2)` Sepration between balls `s=s_(2)-s_(1)=45 t` The first ball strikes the ground. `s_(1)=0=400+10t-1/2 g t^(2)=400+10 t-5 t^(2)` `t^(2)-2t+80=0` `(t-10)(t+8)=0impliest=10s,-8 s` TIME cannot be negative, so `t=10 s`. The second ball strikes the ground. `s_(2)=0=400+55 t-1/2 g t^(2)=400+55 t-5 t^(2)` `t^(2)-11t+80=0` `(t-16)(t+5)=0 implies t=16s` from `t=0` to `t=10 s`, `s=s_(2)-s_(1)=45t` Shape: `y=45x` (inclined straight LINE with +ve slope) From `t= 10` to `16 s, s=s_(2)-0=400+55t-5 t^(2)` Shape:`y=400+55x-5x^(2)` (parabola, open downward)
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