Question

A ball is projected so as to pass a wall at a distance a from the point of projection at an angle of 45° and Falls at a distance be on the other side of the wall. If h is height of the wall then

h=a√2
h=b√2
h=√2ab/a+b
h=ab/a+b

Answer 1

Hope the attachment provided helps you!

Answer 2

Answer:

The height of the wall is an option (d) $h=\frac{ab}{a+b}$.

Explanation:

Calculate the height:

The equation of the trajectory is,

$y=x \tan \theta( \frac{1-x}{R} )\left\rightarrow (1)$

We have, $\theta=45^{\circ}$

Consider, $x=a, R=a+b\right$  

Substitute the values in equation (1)

 $y=a tan45\left(1-\frac{a}{(a+b)}\right)$

The value of $tan 45=1$

So, the equation becomes:

$y=a(1)\left(1-\frac{a}{(a+b)}\right)$

Simplify the equation:

$y=a \frac{a+b-a}{a+b}$

Simplify the numerator:

$y=a \frac{b}{a+b}$

Combine $a$ and $\frac{b}{a+b}:$

$y=\frac{ab}{a+b}$

Hence, the height of the wall is $y=\frac{ab}{a+b}.$