A ball is projected from ground with a speed of 20,m / s at an angle of 45^{circ} with horizontal. T

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3-2.Motion in Plane

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A ball is projected from ground with a speed of $20\,m / s$ at an angle of $45^{\circ}$ with 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 $.........\,m$

A

$5$

B

$7.5$

C

$10$

D

$12.5$

Solution

(b)

$t=\frac{10}{20 \cos 45^{\circ}}=\frac{1}{\sqrt{2}}\,s$

Now, $\quad y=\left(20 \sin 45^{\circ}\right) t-\frac{1}{2} g t^2$

$=20 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}-\frac{1}{2} \times 10 \frac{1}{2}$

$=7.5\,m$

Standard 11

Physics

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