Suppose that instead of the launcher in the lab, you use a much faster one, with

Question

Suppose that instead of the launcher in the lab, you use a much faster one, with a muzzle velocity of 150 m/s. The launcher is tilted at an angle of 38° relative to the ground, and a target is located some distance away, at an elevation 57 m BELOW the launcher. Calculate the following quantities: The initial horizontal and vertical velocities of the projectile. The horizontal distance xr that the launcher should be positioned relative to the target in order for the projectile to hit the target. Use Eq. (3.19). The "hang time" of the projectile before it hits the target. The projectile's horizontal, vertical, and total velocity upon hitting the target.

Explanation / Answer

From the given question,

muzzle velocity=150 m/s

angle of projection=38 degrees

*Initial horizontal velocity=150 cos38=118.2 m/s

Initial vertical velocity= 150 sin38= 92.35 m/s

*Horizontal distance of projectile= 118.2*t

*let Hang time of projectile be t

-57=92.35 t - (1/2)*9.8*t^2

solving we get,

t=19.4 s

Horizontal distance=118.2*19.4=2293.1m

*on hitting the target

Horizontal velocity=118.2 m/s

vertical velocity= 92.35-9.8*19.4=-97.77 m/s

total velocity= sqrt(118.2^2+97.77^2)=153.4 m/s

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

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