A ball at point ' O ' is at a horizontal distance of 7 m from a wall. On the wall a target is set at point ' C '. If the ball is thrown from ' O ' at an angle $37^{\circ}$ with horizontal aiming the target ' $C$ '. But it hits the wall at point ' $D$ ' which is at a vertical distance ' $y_0$ ' below C. If the initial velocity of the ball is $15 \mathrm{~ms}^{-1}$. Find $y_0\left(\right.$ given $\left.\cos 37^{\circ}=\frac{4}{5}\right)$

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Aicharya · Accepted Answer

From figure, $\begin{aligned} & 	an 37^{\circ}=\frac{\mathrm{BC}}{7} \ & 	herefore \mathrm{BC}=7 	imes \frac{3}{4}=\frac{21}{4} \mathrm{~m}=5.25 \mathrm{~m} \ & \mathrm{~V}_{\mathrm{x}}=15 \cos 37^{\circ}=15 	imes \frac{4}{5} \ & =12 \mathrm{~m} / \mathrm{s} \ & 	herefore \mathrm{~S}=\mathrm{v}_{\mathrm{x}} \cdot \mathrm{t} \Rightarrow 7=12 	imes \mathrm{t} \Rightarrow \mathrm{t}=\frac{7}{12} \mathrm{~s} \end{aligned}$ In y -direction, $\begin{aligned} & \mathrm{BD}=\mathrm{V}_{\mathrm{y}} \mathrm{t}-\frac{1}{2} \mathrm{gt}^2=9 	imes \frac{7}{12}-\frac{1}{2} 	imes 10 	imes \frac{7}{12} 	imes \frac{7}{12} \ & =3.54 \mathrm{~m} \ & 	herefore \quad \mathrm{y}_0=\mathrm{BC}-\mathrm{BD}=5.25-3.54=1.71 \mathrm{~m} \end{aligned}$

A ball at point ' O ' is at a horizontal distance of 7 m from a wall. On the wall a target is set at point '…

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