A cannon and a target are `5.10 km` apart and located at the same level. How soon will the shell launched with the initial velocity `240 m//s` reach the target in the absence of air drag?

Answer

`0.41` or `0.71min` later, depending on the initial angle.

Explanation

Total time of motion
`tau=(2v_0sin alpha)/(g)` or `sin alpha=(taug)/(2v_0)=(9*8tau)/(2xx240)` (1)
and horizontal range
`R=v_0cos alpha tau` or `cos alpha=(R)/(v_0tau)=(5100)/(240tau)=(85)/(4tau)` (2)
From Eqs. (1) and (2)
`((9*8)^2tau^2)/((480)^2)+((85)^2)/((4tau^2)^2)=1`
On simplifying `tau^4-2400tau^2+1083750=0`
Solving for `tau^2` we get:
`tau^2=(2400+-sqrt(1425000))/(2)=(2400+-1194)/(2)`
Thus `tau=42.39s=0.71min` and
`tau=24.55s=0.41min` depending on the angle `alpha`.