A battleship simultaneously fires two shells toward two identical enemy ships. O

Question

A battleship simultaneously fires two shells toward two identical enemy ships. One shell hits ship A, which is close by, and the other hits ship B, which is farther away. The two shells are fired at the same speed. Assume that air resistance is negligible and that the magnitude of the acceleration due to gravity is .

For two shells fired at the same speed which statement about the horizontal distance traveled is correct?
The shell fired at a larger angle with respect to the horizontal lands farther away.

The shell fired at an angle closest to 45 degrees lands farther away.

The shell fired at a smaller angle with respect to the horizontal lands farther away.

The lighter shell lands farther away.

Explanation / Answer

According to projectile motion, the range of the projectile is R = V^2sin2/2g the above equation uses 2*, 45 is the angle that gives the highest answer. If the angle is 44 degrees, sin(2*44) = .999, and sin(2*46) = .999: however, sin(2*45)= 1
Basically, 45 is the angle at which the speed of the projectile isevenly split between horizontal and vertical components.

So option B is correct. B) the shell fired at an angle closest to 45 degreeslands farther away

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.