QUESTION

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# Danielle launches a ball off of a platform of height ℎh with velocity ?⃗ 0v→0 at an angle ?θ above the horizontal. A wall with a ball‑sized hole a height ?H above the floor, with ?>ℎ,H>h, is a horizontal distance ?=11.0 mL=11.0 m away from Danielle. The difference in height between ?H and ℎh is one‑fifth of ?,L, and the square of the magnitude of the velocity, ?20,v02, is five‑fourths of the product ??.gL.?−ℎ?20=15?=54??H−h=15Lv02=54gLDetermine the two values of ?θ that ensure the ball passes through the hole. Submit the larger angle as ?1θ1 and the smaller angle as ?2.Danielle launches a ball off of a platform of height $h$ with velocity $\vec{v}_{0}$ at an angle $\theta$ above the horizontal. A wall with a ball-sized hole a height $H$ above the floor, with $H>h$, is a horizontal distance $L=11.0 \mathrm{~m}$ away from Danielle. The difference in height between $H$ and $h$ is one-fifth of $L$, and the square of the magnitude of the velocity, $v_{0}^{2}$, is five-fourths of the product $g L$. $\begin{array}{r} H-h=\frac{1}{5} L \\ v_{0}^{2}=\frac{5}{4} g L \end{array}$ Determine the two values of $\theta$ that ensure the ball passes through the hole. Submit the larger angle as $\theta_{1}$ and the smaller angle as $\theta_{2}$. $\theta_{1}=$ $\theta_{2}=$  