Question

A projectiel is fired from a gun has an initial velocityof

90.0 Km/h at an angle of 60.0 to the horizentale. when
theprojectile is at the top of its trajectory an internal
explosioncauses it to separate into two fragments that have equal
masse, oneof the fragments falls straight downward as though it has
beenreleased from rest. How far horizentally does the second piece
landfrom the gun?

Answer 1

Step-by-step explanation:

Given

Launch velocity
`u=90 km/h\approx 25 m/s`

inclination
`\theta =60^(\circ)`

At highest point the projectile exploded such that one part falls vertically downward and other part travel in a trajectory such that it land on earth at distance of x m from original position of center of mass

As the force is internal so center of mass of the system will remain at the same position. So particle will land at such spots such that com will remain at Range of Projectile.

Original Position of center mass is given by the Range of Projectile

`Range=(u^2\sin 2\theta )/(g)`

`R=(25^2\sin 2(60))/(9.8)`

`R=55.23\ m`

at Highest position distance of first part from center of mass will be
`=R-(R)/(2)=(R)/(2)`

product of mass and distance from center of mass should be same therefore

`(m)/(2)* (R)/(2)=(m)/(2)* x`

thus
`x=(R)/(2)`

i.e. Second part is at a distance of
`(R)/(2)` from center of mass

From gun it is
`R+(R)/(2)=(3R)/(2)=(3* 55.23)/(2)=82.84 m`