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 leve

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» Correct Answer - `0.41` or `0.71min` later, depending on the initial angle. Total time of motion `tau=(2v_0sin alpha)/(g)` or `sin alpha=(taug)/(2v_0)=(98tau)/(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) `((98)^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`.

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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. How soon will the shell launched with the initial velocity `240 m//s` reach the target in the absence of air drag?

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