From the top of tower of height $80 m$ , two stones are projected horizontally with $20 m / s e c$ and $30 m / s e c$ in opposite directions. Find the distance between both the stones on reaching the ground.

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Solution

We know
$s = u t + \frac{1}{2} g t^{2}$
here $u = 0$
$g = 10 m / s^{2}$
$s = 80 m$
$80 = \frac{1}{2} \times 10 t^{2}$
$80 = 5 t^{2}$
$t^{2} = 16$
$t = 4 s e c$
For stone 1, $d_{1} = v t = 20 \times 4 = 80 m$
For stone 2, $d_{2} = v t = 30 \times 4 = 120 m$
Total distance $D = d_{1} + d_{2}$
$= 120 + 80 = 200 m$

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From the top of tower of height 80m, two stones are projected horizontally with 20m/sec and 30m/sec in opposite directions. Find the distance between both the stones on reaching the ground.

We know s=ut+12gt2here u=0g=10 m/s2s=80 m80=12×10t280=5t2t2=16t=4 secFor stone 1, d1=vt=20×4=80 mFor stone 2, d2=vt=30×4=120 mTotal distance D=d1+d2 =120+80=200 m

From the top of tower of height $80 m$ , two stones are projected horizontally with $20 m / s e c$ and $30 m / s e c$ in opposite directions. Find the distance between both the stones on reaching the ground.