From the ground level,a ball is to be shot wi | vedclass

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(B) The range of a projectile is given by $R = \frac{u^2 \sin(2	heta)}{g}$.From the graph,the maximum range $R_{max} = 250 \ m$ occurs at $	heta = 4

Vedclass · Accepted Answer

$26.5^o$ and $63.5^o$

Vedclass · Answer

(B) The range of a projectile is given by $R = \frac{u^2 \sin(2	heta)}{g}$.From the graph,the maximum range $R_{max} = 250 \ m$ occurs at $	heta = 45^o$.Thus,$R_{max} = \frac{u^2}{g} = 250 \ m$.For a given range $R = 200 \ m$,we have $200 = 250 \sin(2	heta)$.$\sin(2	heta) = \frac{200}{250} = 0.8$.$2	heta = \arcsin(0.8)$.The two possible values for $2	heta$ are $2	heta_1 = 53.13^o$ and $2	heta_2 = 180^o - 53.13^o = 126.87^o$.Therefore,$	heta_1 = \frac{53.13^o}{2} \approx 26.5^o$ and $	heta_2 = \frac{126.87^o}{2} \approx 63.5^o$.

From the ground level,a ball is to be shot with a certain speed. The graph shows the range $(R)$ of the particle versus the angle of projection from the horizontal $(\theta)$. The values of $\theta_1$ and $\theta_2$ are:

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