Algebra 2 (1st Edition)

$\approx 10.05$ ft more
Here, we have $\theta_1=45^{\circ}$ and $\theta_2=60^{\circ}$ We know that $d=\dfrac{v^2}{32} \sin (2 \theta)$ $d=\dfrac{49^2}{32} \sin 2 (45^{\circ}) \approx 75.03$ft ...(1) and $d=\dfrac{49^2}{32} \sin 2 (60^{\circ}) \approx 64.98$ft ...(2) Now, we get $75.03-64.98 \approx 10.05$ ft more