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(B) The horizontal component of velocity is $u_x = u \cos 	heta = 20 \cos 45^{\circ} = 20 	imes \frac{1}{\sqrt{2}} = 10\sqrt{2}\,m/s$.The vertical c

Vedclass · Accepted Answer

$7.5$

Vedclass · Answer

(B) The horizontal component of velocity is $u_x = u \cos 	heta = 20 \cos 45^{\circ} = 20 	imes \frac{1}{\sqrt{2}} = 10\sqrt{2}\,m/s$.The vertical component of velocity is $u_y = u \sin 	heta = 20 \sin 45^{\circ} = 20 	imes \frac{1}{\sqrt{2}} = 10\sqrt{2}\,m/s$.The time $t$ taken to reach the wall at a distance $x = 10\,m$ is given by $t = \frac{x}{u_x} = \frac{10}{10\sqrt{2}} = \frac{1}{\sqrt{2}}\,s$.The height $y$ at which the ball hits the wall is given by the equation of trajectory: $y = u_y t - \frac{1}{2} g t^2$.Substituting the values (taking $g = 10\,m/s^2$): $y = (10\sqrt{2}) 	imes \frac{1}{\sqrt{2}} - \frac{1}{2} 	imes 10 	imes (\frac{1}{\sqrt{2}})^2$.$y = 10 - 5 	imes \frac{1}{2} = 10 - 2.5 = 7.5\,m$.

$A$ ball is projected from the ground with a speed of $20\,m/s$ at an angle of $45^{\circ}$ with the horizontal. There is a wall of $25\,m$ height at a distance of $10\,m$ from the projection point. The ball will hit the wall at a height of $.........\,m$.

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