A cannon shoots a rock from the ground at an angle of 31.0° above the horizontal

Question

A cannon shoots a rock from the ground at an angle of 31.0° above the horizontal. The rock's initial speed is 204.0 m/s. (Neglect the height of the shooting point and air resistance.)
(a) What is the rock's speed in the horizontal direction just before it hits the ground?

I'm able to figure out a final velocity given that the initial shooting point is either higher or lower than the final landing point, but for whatever reason cannot figure this problem out. Any help clearing this up is much appreciated.

Explanation / Answer

We do the whole projectile motion by resolving the velocity along two directions,

horizontal --------vcos

vertical ----------vsin

As there is no gravity involved ,,when the projectile is covering hporizontal motion, so no acceleration

if there is no accelertaion or retardation, final velocity will remain the initial velocity

So the horizontal velocity of rock will remain as 204 cos31 in the entire path

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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