laws of motion

a body is projected at an angle $45^o$ to the horizontal has a range of 16m. it explodes into two equal parts at the highest point out of which one falls vertically downward from the point of projection while the other moves horizontally. then what is the distance at which the latter will fall from the point of projection? the answer is 24m.

the way i tried solvin:
$R=\frac{u^2 sin2\theta}{g}[br]so, u=\sqrt{160}[br]$

now at highest point,$\frac{mu(cos\theta)^2}{2}= \frac{(m/2)v^2}{2}[br]$
so, v=u
now from $h=\frac{gt^2}{2}$ and $s=vt$
we get,
$s=4 \sqrt{8}$
then we must have required distance = 16/2 +$4 \sqrt{8}$
but the answers iS 24.
what's wrong here?

Your mistake is the equation s=ut

It is instead s=u*(sin(theta))*t= (ut)*sin(theta)=(4(8)^0.5)*(2^(0.5))=4*4=16

Hence, s=16/2+16=24

You should use conservation of momentum because there is no external force acting on the ball in the horizontal direction ,

According to which the momentum in the horizontal direction is conserved .

The net momentum before collision is m*u*cos(theta)

[m]

After collision the momentum of one of the ball in horizontal direction is 0. And that of be (m/2)*v

Where v is the velocity of half-ball in horizontal direction.

Hence, (m/2)*v= m*u*(2^(-0.5))

v=(2^(0.5))*u

Note that ball has already reached some height and will take u/g time to reach the ground, In this time the ball will travel v*(u*sin(theta)/g) distance in horizontal direction, which will be equal to u^2/g= (160)/10= 16.

Hence the ball will fall at distance of 16/2 +16=24 meter away from the point of projection

why is $u*sin\theta$ used instead of u.

thank you for the help


i think i found my mistake. i shouldn't have written cos(theta) right? i realized after i saw that you had use the conservation of momentum. we can use conservation of kinetic energy too right?