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  • A ball projected from ground at an angle of  just clears a wall in front. If point of proj

A ball projected from ground at an angle of 45^{\circ} just clears a wall in front. If point of projection is 4m from the foot of wall and ball strikes the ground at a distance of 6m on the other side of the wall, the height of the wall is : 

Option: 1

4.4m


Option: 2

2.4m


Option: 3

3.6m


Option: 4

1.6m


Answers (1)

best_answer

                                   
As the ball is projected at an angle 45o

to the horizontal therefore Range = 4H

10=4 H \Rightarrow H=\frac{10}{4}=2.5 m (\text {Range}=4 m +6 m =10 m )

\\ H=\frac{u^{2} \sin ^{2} \theta}{2 g} \therefore u^{2}=\frac{H \times 2 g}{\sin ^{2} \theta}=\frac{2.5 \times 2 \times 10}{\left(\frac{1}{\sqrt{2}}\right)^{2}}=100 \ or ,\\ \quad u=\sqrt{100}=10 ms ^{-1}

 

 Height of wall PA = 

O A \tan \theta-\frac{1}{2} \frac{g(O A)^{2}}{u^{2} \cos ^{2} \theta}=4-\frac{1}{2} \times \frac{10 \times 16}{10 \times 10 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}}=2.4 m

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Rishi

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