Let the horizontal distance between the point of drop and the net be 'R' metres and it takes 't' time to reach
the y component of initial velocity is zero {uy=0; u=ux=120 m/s}
given,
h=100 m ; u=120 m/s;
a) for vertical motion (accelerated)
h=uyt +0.5gt2
100=0 + 0.5*9.8*t2
t = 4.517 seconds
b) for horizontal motion (unaccelerated)
R=t*ux = 4.517*120 = 542.04 m
c) since the horizontal motion (in x direction) is unaccelerated hence, the final velocity in x direction remains unchanged.
vx=ux=120 m/s
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