Shots are fired from the top of a tower and f | vedclass

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(A) Let the point of collision be at a horizontal distance $a$ from the tower and at a height $y$ from the ground.For the projectile fired from the bo

Vedclass · Accepted Answer

$\frac{2a}{\sqrt{3}}$

Vedclass · Answer

(A) Let the point of collision be at a horizontal distance $a$ from the tower and at a height $y$ from the ground.For the projectile fired from the bottom with velocity $u_2$ at $60^o$:$a = u_2 \cos 60^o \cdot t$$y = u_2 \sin 60^o \cdot t - \frac{1}{2}gt^2 = a 	an 60^o - \frac{1}{2}gt^2$For the projectile fired from the top with velocity $u_1$ at $30^o$:$a = u_1 \cos 30^o \cdot t$The vertical position relative to the ground is $y = h + u_1 \sin 30^o \cdot t - \frac{1}{2}gt^2 = h + a 	an 30^o - \frac{1}{2}gt^2$Equating the two expressions for $y$:$a 	an 60^o - \frac{1}{2}gt^2 = h + a 	an 30^o - \frac{1}{2}gt^2$$h = a(	an 60^o - 	an 30^o)$$h = a(\sqrt{3} - \frac{1}{\sqrt{3}}) = a(\frac{3-1}{\sqrt{3}}) = \frac{2a}{\sqrt{3}}$

Shots are fired from the top of a tower and from its bottom simultaneously at angles $30^o$ and $60^o$ as shown. If the horizontal distance of the point of collision is at a distance $a$ from the tower,then the height of the tower $h$ is:

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