Two identical shells are fired from a point on the ground with same muzzle velocity at angle of elevation `alpha=45^(@)` and `beta=tan^(-1)3` towards top of a cliff, `20 m` away from the point of firing. If both the shells reach the top simultaneously, then
height of the cliff is

Options

A. `20m`
B. `10 m`
C. `15m `
D. `30m`

Answer

B

Explanation

Let muzzles velocity be `u_(0)` and height of cliff be `h`. Assuming horizontal direction to be positive `x`-axis and vertically upward direction to be positive `y`-axis, coordinates of top of cliff become `(20, h)` as shown in the figure.

Using equation of trajectory of a projectile for two shells,
`h=20tanalpha-(g(20)^(2))/(2u_(0)^(2)cos^(2)alpha)`
`=20tanbeta-(g(20)^(2))/(2u_(0)^(2)cos^(2)beta)`
From the two equations
`u_(0)=20 m//s` and `h=10 m `
Time taken by the shell, fired at angle `alpha` to reach to the top of cliff is
`t_(1)=20/(u_(0)cosalpha)=sqrt(2)s`
Time taken by the other shell is
`t_(2)=20/(u_(0)cosbeta)=sqrt(10)s`
Hence, the shell having angle of projection `beta` was fired first and other shell (having angle of projection `alpha`) was fired later.
Time interval between two firings is
`t_(2)-t_(1)=(sqrt(10)-sqrt(2))s=1.74s`
Consider vertical velocities of shells just before striking the top of cliff.
For the first shell,
`v_(1y)=u_(0)sinalpha-"gt"_(1)=20xx1/sqrt(2)-10sqrt(2)=0`
For the second shell,
`v_(2y)=u_(0)sinbeta-"gt"_(2)=20xx3/sqrt(10)-10sqrt(10)=-4sqrt(10)m//s`
Let `v_(y)` be combined vertical velocity of the shell after sticking together, then from conservation of momentum in vertical direction, we get
`(m+m)v_(Y)=mxx0+m(-4sqrt(10))`
`implies v_(y)=-sqrt(10) m//s`
Collision with the top of cliff is perfectly elastic, so combined shell will rebound with velocity `2sqrt(10) m//s` in vertical direction. Maximum height above top of cliff will be `(2sqrt(10))^(2)//2g=2m`.
Net maximum height `=10+2=12m`