A body projected at an angle 45° with horizontal has range 16 m. It explodes into two parts of equal masses at the highest point. One of parts falls downwards at the point of explosion. At what distance from the point of throw, the other will fall ?

Options

A. 8m
B. 16m
C. 24m
D. 32m

Answer

D

Explanation

Here `theta=45^@`. The range is maximum = `u^(2)/g=16m`
`:.u=4sqrt(10)ms^(-1)`
Applying law of conservation of momentum at the highest point.
`mucostheta=m/2(-ucostheta)+m/2v`
This gives `ucostheta=(-ucostheta)/2+v/2`
`3ucostheta=v`…(i)
or `v=3xx4sqrt(10)xx1/sqrt(2)=12sqrt(5)ms^(-1)`
Max. height, `h=R_(max)/4=4m`
Time taken for vertical fall `t=sqrt((2h)/g)=sqrt((2xx4)/10)=2/sqrt(5)s`
`:.` Horizontal distance covered
`x=vxxt=12sqrt(5)xx2/sqrt(5)=24m`
Total distance=R/2+x
Total distance = 8 + 24 = 32 m from point of throw.