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?

A cannon and a target are 5.10 km apart and located at the same level.

1.	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» Solution :Hence `v_(0)=240ms^(-1),R=5.10ikm=5100m,,g=9.8ms^(-2), theta=?` From formula `R=(v_(0)^(2)sin 2 theta)/g` We have `sin 2 theta=(Rg)/(v_(0)^(2))`, Putting VALUES we get `sin 2 theta =(5100xx9.8)/(240xx240)=0.8677=(sqrt(3))/2=sin60^(@)` or `sin 120^(@)impliestheta=30^(@)` or `60^(@)` From formula `T=(2v_(0)sin theta)/g` When `theta =30^(@),T_(1)=(2xx240xx0.5)/9.8=24.5`sec When `theta=60^(@),T_(2)=(2xxd240xx0.867)/9.8=42.14` sec