A body of mass $$m$$ is projected with a speed $$u$$ making an angle of $$45°$$ with the ground. The angular momentum of the body about the point of projection, at the highest point is expressed as $$\frac{\sqrt{2}mu^3}{Xg}$$. The value of $$X$$ is

given
mass = m
projection speed = u
angle = 45°

we need angular momentum about point of projection at highest point

formula:

L = m | r × v |

for 2D motion:

$$L=m(xv_y−yv_x)$$

step 1: velocity at highest point

at highest point
vertical velocity = 0
horizontal velocity =

$$u\cos45°=u/\sqrt{2}$$

so:

v_x = u/√2 , v_y = 0

step 2: time to reach highest point

$$t=(u\sin45°)/g$$

= (u/√2)/g

= u/(√2 g)

step 3: coordinates of highest point

horizontal distance:

$$x=(u\cos45°)\times t$$

= (u/√2) × (u/(√2 g))

= u²/(2g)

vertical height:

$$y=(u\sin45°)^2/(2g)$$

= (u²/2)/(2g)

= u²/(4g)

step 4: angular momentum

$$L=m(xv_y−yv_x)$$

$$\sin ce\ v_y=0:$$

$$L=−myv_x$$

magnitude:

$$L=myv_x$$

$$=m\times(u^2/(4g))\times(u/\sqrt{2})$$

$$=mu^3/(4\sqrt{2}g)$$

Now comparing with the given expression :-

$$\frac{\sqrt{2}mu^3}{Xg}$$

we get it by multiplying and dividing our answer by $$\sqrt{\ 2}$$

so we get $$=\sqrt{\ 2}mu^3/8g$$
so Value of X is 8