Two identical shells are fired from a point on the ground with same muzzle velocity at angle of elevation  and  towards top of a cliff, 20 m away from the point of firing. If both the shells reach the top simultaneously, then

Muzzle velocity is
a)	10 m/s	b)	5 m/s	c)	15 m/s	d)	20 m/s

(d)

Let muzzle velocity be  and height of cliff be . Assuminghorizontal direction to be positive -axis and vertically upward direction to be positive -axis, coordinates of top of cliff become  as shown in the figure

Using equation of trajectory of a projectile for two shells,

From the above two equations,

and

Time taken by the shell, fired at angle , to reach the top of cliff is

Time taken by the shell is

Hence, the shell having angle of projection  was fired first and the other shell (having angle of projection ) was fired later

Time interval between two firings is

Consider vertical velocities of shells just before striking the top of cliff

For the first shell,

For the second shell,

Let be combined vertical velocity of the shell after sticking together, then from conservation of momentum in vertical direction, we get

 m/s

Collision with the top of cliff is perfectly elastic, so combined shell will rebound with velocity  m/s in vertical direction. Maximum height above top of cliff will be

Net maximum height = 10 + 2 = 12 m