(B) The maximum range of a projectile is achieved when the angle of projection is $\theta = 45^{\circ}$.The formula for the range $R$ is given by $R =

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(B) The maximum range of a projectile is achieved when the angle of projection is $\theta = 45^{\circ}$.The formula for the range $R$ is given by $R = \frac{u^2 \sin(2\theta)}{g}$.For maximum range, $\sin(2\theta) = \sin(90^{\circ}) = 1$.Thus, $R_{max} = \frac{u^2}{g}$.Given $u = 30 \,m/s$ and $g = 10 \,m/s^2$, we have:$R_{max} = \frac{(30)^2}{10} = \frac{900}{10} = 90 \,m$.

Class 11 Physics 3-2.Motion in Plane Horizontal Projectile Motion

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$A$ cricket player can throw a ball with an initial speed of $30 \,m/s$. What is the maximum range the player can throw the ball? Neglect air resistance. [Take $g = 10 \,m/s^2$]

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A
$100 \,m$
B
$90 \,m$
C
$80 \,m$
D
$90\sqrt{2} \,m$

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3-2.Motion in Plane

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Horizontal Projectile Motion

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$A$ cricket player can throw a ball with an initial speed of $30 \,m/s$. What is the maximum range the player can throw the ball? Neglect air resistance. [Take $g = 10 \,m/s^2$]

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