A dart is thrown horizontally toward X at 25 ms^-1 as shown in figure. It hits Y 0.12 s later. The distance XY is?

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1 Answer
Jun 27, 2017

s_(XY)=0.07 m (=7 cm)

Explanation:

Because we can consider the horizontal and vertical directions independently, this is actually exactly the same question as 'how far will an object, dropped from rest, fall in 0.12 s?'

The initial velocity in the horizontal, x, direction u_x=25 ms^-1. This will stay constant, but we can ignore it.

The initial velocity in the vertical, y, direction is u_y=0 ms^-1. Its acceleration will simply be the acceleration due to gravity, a_y=9.8 ms^-2

s_(XY)=u_yt+1/2a_yt^2=0xx0.12+1/2xx9.8xx0.12^2

s_(XY)=0.07 m (=7 cm)