021 (part 1 of 7) 10.0 points A projectile of mass 0.778 kg is shot from a canno

Question

021 (part 1 of 7) 10.0 points A projectile of mass 0.778 kg is shot from a cannon. The end of the cannon's barrel is at height 6.6 m, as shown in the figure. The initial velocity of the projectile is 10 m/s The projectile rises to a maximum height of above the end of the cannon's barrel and strikes the ground a horizontal distance past the end of the cannon's barrel. Determine the vertical component of the initial velocity at the end of the cannon's bar- rel, where the projectile began its trajectory The acceleration of gravity is 9.8 m/s Answer in units of m/s

Explanation / Answer

21. v0y = 10 sin49 = 7.55 m/s

22. at maximum height, vy = 0

applying vf^2 - vi^2 = 2 a d

0^2 - 7.55^2 = 2(-9.8)(delta(y))

delta(y) = 2.91 m

23. vf = vi + a t

0 = 7.5 - 9.8 t

t = 0.77 sec

24. in vertical,

yf - yi = v0y t + ay t^2 /2

0 - (6.6 sin49)= 7.55t - 9.8t^2/2

4.9 t^2 - 7.55t - 5 = 0

t = 2.04 s

25. vx = v0x = 10 cos49 = 6.56 m/s

vy = v0y + ay t = (7.55) + (-9.8 x 2.04)

vy = -12.4 m/s

magnitude = sqrt(vx^2 + vy^2)

= 14 m/s

26. theta = tan^-1(-12.4 / 6.56) = - 62 deg

Ans: 62 deg

27. delta(x) = (v0x)t

= 6.56 x 2.04

= 13.4 m

Related Questions

Navigate

• Browse (All)
• Browse
• Subjects

---

---

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