Question

An arche fires an arrow horizontally with a speed of 89m/s directly at the bull's eye of a target 80 metes away.
The arrow, at thr tim it was shot, and the bull;s eye are both 1 meter above the ground.
How fa short of the target does the arow strike the ground?

Answer 1

Okay! So first why don't you try to find out how far away the arrow lands from the archer!
Well the time in air can be calculated by using this formula!
\[\Delta y = v_{yi} \Delta t + \frac{a \Delta t^{2}}{2}\]
But the arrow has no initial vertical component, so that drops out! We know how high it started, and we know it lands at the ground!
\[2(y_{f}-y_{i}) = a \Delta t^{2}\]
yf is the final position (0) and yi is the intial position 1
\[\sqrt{\frac{2(0 -1)}{a}}= t \]
And we know a is -9.8
\[\sqrt{\frac{-2}{-9.8}} = t = .45s\]
Okay....then we know that the horizontal distance traveled, is equal to the velocity multiplied by the time in air.
\[x = v_{x} \Delta t\]
Then just plug in the quantities
So it would fall short.... 80 - x