The ceiling of a long hall is 25; m high. Wha | vedclass

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(A) Given: Speed of the ball,$u = 40\; m/s$. Maximum height,$h = 25\; m$. Acceleration due to gravity,$g = 9.8\; m/s^2$.In projectile motion,the maxim

Vedclass · Accepted Answer

$150.53$

Vedclass · Answer

(A) Given: Speed of the ball,$u = 40\; m/s$. Maximum height,$h = 25\; m$. Acceleration due to gravity,$g = 9.8\; m/s^2$.In projectile motion,the maximum height reached by a body projected at an angle $	heta$ is given by the relation:$h = \frac{u^2 \sin^2 	heta}{2g}$Substituting the values:$25 = \frac{(40)^2 \sin^2 	heta}{2 	imes 9.8}$$25 = \frac{1600 \sin^2 	heta}{19.6}$$\sin^2 	heta = \frac{25 	imes 19.6}{1600} = 0.30625$$\sin 	heta = \sqrt{0.30625} = 0.5534$$	heta = \sin^{-1}(0.5534) = 33.60^{\circ}$The horizontal range $R$ is given by:$R = \frac{u^2 \sin(2	heta)}{g}$$R = \frac{(40)^2 \sin(2 	imes 33.60^{\circ})}{9.8}$$R = \frac{1600 	imes \sin(67.2^{\circ})}{9.8}$$R = \frac{1600 	imes 0.9219}{9.8} \approx 150.53\; m$.

The ceiling of a long hall is $25\; m$ high. What is the maximum horizontal distance that a ball thrown with a speed of $40\; m/s$ can go without hitting the ceiling of the hall (in $; m$)?

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