Answer:
The dart won't hit within 1.00 cm of the bullseye
Explanation:
First, we need to calculate the time that the dart takes to hit the dartboard. This can be calculated using the distance of 5.7 m and the horizontal speed of 21.3 m/s, so
[tex]\begin{gathered} v=\frac{d}{t} \\ \\ t=\frac{d}{v}=\frac{5.7\text{ m}}{21.3\text{ m/s}}=0.268\text{ s} \end{gathered}[/tex]
Then, with this time we can calculate the change in height of the dart, using the following equation
[tex]\Delta y=v_0t-\frac{1}{2}gt^2[/tex]
Where vo is the initial vertical velocity, so vo = 0 m/s, g is the gravity, so g = 9.8 m/s², and t is 0.268 s.
So, replacing the values, we get:
[tex]\begin{gathered} \Delta y=0(0.268)-\frac{1}{2}(9.8)(0.268)^2 \\ \Delta y=0.3509\text{ m} \\ \Delta y=35.09\text{ cm} \end{gathered}[/tex]
Since Δy is greater than 1 cm, the dart will not hit within 1.00 cm of the bullseye. So, to adjust the shot, you should release the dart from 35.09 cm above.
