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

Question

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

Smart Achievers · Accepted Answer

(2)

As ball is projected at an angle $45 °$ to the horizontal therefore Range $= 4 H$

or $10 = 4 H \Rightarrow H = \frac{10}{4} = 2.5 m$

Height of wall $P A = O A \tan θ - \frac{1}{2} \frac{g {(O A)}^{2}}{u^{2} \cos^{2} θ}$

$= 4 - \frac{1}{2} \times \frac{10 \times 16}{10 \times 10 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}} = 2.4 m$

Solution

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

1 4.4 m

2 2.4 m

3 3.6 m

4 1.6 m

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