A ball is shot from the ground into the air. At a height of 8.2 m, the velocity

Question

A ball is shot from the ground into the air. At a height of 8.2 m, the velocity is observed to be

v = (7.4)i + (5.3)j in meters per second (i horizontal, j upward).

a. To what maximum height will the ball rise?

b.What will be the total horizontal distance traveled by the ball?

c. What is the velocity of the ball (magnitude and direction) the instant before it hits the ground? That is what is its i-component?

d. What is its j-component?

Explanation / Answer

Ignoring air resistance, the horizontal velocity is constant at 6.7m/s Vy(t) = Vyi -g*t =>Vy(t1) = 6 = Vyi -9.8*t1 => Vyi = 6+9.8*t1 h(t1) = 9.1 = Vyi*t1-4.9*t1^2 = 6*t1+9.8*t1^2-4.9*t1^2 4.9t1^2+6*t1-9.1=0 Using the quadratic formula, t1 = 0.88174s, substitute into the equation above for Vyi: Vyi = 14.64m/s At the moment of peak altitude, the vertical velocity slows to a stop before falling at g. Vy(t) = Vyi-g*t = 0 => t =14.64/g = 1.494 at max height h(1.494) = 14.64*1.494 - 4.9*1.494^2 = 10.94m a)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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