A projectile is launched at speed v0 = 30\text{ m/s} at \theta = 35° above level ground. Using g = 9.8\text{ m/s}^2, what is the horizontal range to the nearest meter?

86 m Range formula on level ground: R = v0^2 \sin(2\theta)/g. Plug in \sin 70° \approx 0.9397. Then R = (900)(0.9397)/9.8 \approx 86.3\text{ m}.

Vectors and Motion in Two Dimensions

AP Physics 1· difficulty 4/5

v_0 = 30\text{ m/s}

Explanation

R = v_0^2 \sin(2\theta)/g

Want 10 more like this — adaptive to your weak spots?