A ball is projected at an angle of 45^{circ} | vedclass

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Question

(B) The range of the projectile is given by $R = d_1 + d_2 = \frac{u^2 \sin(2	heta)}{g}$.Since $	heta = 45^{\circ}$,$\sin(2 	imes 45^{\circ}) = \si

Vedclass · Accepted Answer

$h=\frac{d_1 d_2}{d_1+d_2}$

Vedclass · Answer

(B) The range of the projectile is given by $R = d_1 + d_2 = \frac{u^2 \sin(2	heta)}{g}$.Since $	heta = 45^{\circ}$,$\sin(2 	imes 45^{\circ}) = \sin(90^{\circ}) = 1$.Thus,$R = d_1 + d_2 = \frac{u^2}{g}$,which implies $\frac{u^2}{g} = d_1 + d_2$.The equation of the trajectory is $y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$.Substituting $y = h$,$x = d_1$,and $	heta = 45^{\circ}$:$h = d_1 	an 45^{\circ} - \frac{g d_1^2}{2u^2 \cos^2 45^{\circ}}$.Since $	an 45^{\circ} = 1$ and $\cos^2 45^{\circ} = \frac{1}{2}$,we have:$h = d_1 - \frac{g d_1^2}{2u^2 (1/2)} = d_1 - \frac{g d_1^2}{u^2}$.Substituting $\frac{u^2}{g} = d_1 + d_2$:$h = d_1 - \frac{d_1^2}{d_1 + d_2} = \frac{d_1(d_1 + d_2) - d_1^2}{d_1 + d_2} = \frac{d_1^2 + d_1 d_2 - d_1^2}{d_1 + d_2} = \frac{d_1 d_2}{d_1 + d_2}$.

$A$ ball is projected at an angle of $45^{\circ}$ with the horizontal. It passes through a wall of height $h$ at a horizontal distance $d_1$ from the point of projection and strikes the ground at a distance $d_1+d_2$ from the point of projection. Then $h$ is:

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