Based off those answers, you have the following.. Example 3.2 That\'s Quite an A

Question

Based off those answers, you have the following..

Example 3.2 That's Quite an Arm! A stone is thrown from the top of a building upward at an angle of 28.0° to the horizontal with an initial speed of 15.6 m/s as shown in the figure. The height from which the stone is thrown is 45.0 m above the ground (A) How long does it take the stone to reach the ground? (B) What is the speed of the stone just before it strikes the ground? 45.0 m SOLVE IT Conceptualize Study the figure, in which we have indicated the trajectory and various parameters of the motion of the stone A stone is thrown from the top of a building Categorize We categorize this problem as a projectile motion problem. The stone is modeled as a particle under constant acceleration in the y direction and a particle under constant velocity in the x direction Analyze we have the information x, = y, = 0, yf=-45.0 m, a,--g, and vi = 15.6 m/s (the numerical value of yf is negative because we have chosen the point of the throw as the origin)

Explanation / Answer

A)

Vxi = vi*cos(theta)
= 15.6*cos28
= 13.77 m/s

Vyi = vi*sin(theta)
= 15.6*sin28
= 7.32 m/s

yf = yi + Vyi*t+ 1/2*ay*t^2

=> -45 = 0 + Vyi*t + 1/2(-9.8)t^2

=> -45 = 7.32t - 4.9t^2

=> t = 3.87 s

B)

Vyf = Vyi + ay*t
=> Vyf = Vyi + (-9.8)t
=> Vyf = 7.32 + (-9.8)(3.87)
=> Vyf = -30.6 m/s

Vf = sqrt(Vyf^2 + Vxf^2)
=> sqrt(-30.6^2 + 13.77^2)
=> Vf = 33.5 m/s

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---

---

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