A shell is fired from a cliff that is 36m above a horizontalplane. The muzzle sp

Question

A shell is fired from a cliff that is 36m above a horizontalplane. The muzzle speed of the shell is 80.0m/s and it is fired atan elevation of 25 degrees above the horizontal. What is the velocity of the shell as it strikes theground? Please show steps, thanks! :) A shell is fired from a cliff that is 36m above a horizontalplane. The muzzle speed of the shell is 80.0m/s and it is fired atan elevation of 25 degrees above the horizontal. What is the velocity of the shell as it strikes theground? Please show steps, thanks! :)

Explanation / Answer

Find the y component of the velocity: -- V(y) = 80Sin V(y) = 80Sin(25) V(y) = 33.8 m/s -- Now, firnd the height when velocity is zero: -- v(f)^2 = v(i)^2 + 2*a*d 0 = 33.8^2 + 2*9.8*d d = 1143 / 19.6 d = 58.3 meters -- Total height used to calculate the velocity is: total height = h(o) + d = 36 m + 58.3 m = 94.3 m -- And the vertical velocity, with conservation of energyis: 1/2mv^2 = mgh v = (2gh) v = 43 m/s -- -- Hope this helps.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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