Here's the question you clicked on:
kbarnes7
a machine can toss a golf ball a distance of 120 m on a level playing field when the machine is set to an angle of 30 degrees. How far can the machine toss the ball when it is set to an angle of 45 degrees.
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ok so where do i go from there
\[R=\frac{u^2 \sin(2 \theta)}{g}\] \[\theta = 30 degrees\]R=120m \[120=\frac{u^2 \sin(2 *30)}{10}\] find u.
then find R for 45 degrees with this u.
ok, so is it 170 m, 98 m , 140 m , 85 m or 150 m? trying to check my answer
what u you got?
\[u^2 = \frac{120*10}{\sin 60}\] \[u^2 = 800\sqrt{3}\] now theta = 45 \[R=\frac{u^2 \sin(2*45)}{10}\] substitute value of u^2. \[R=\frac{800 \sqrt{3} \sin(2*45)}{10}\]
\[R=80\sqrt{3}\]
man i missed that up, so with sig figs.... 140 m?