12-6. A ba s released from the bottom of an elevator which is treling upward wit

Question

12-6. A ba s released from the bottom of an elevator which is treling upward with a velocity of 6ft/s. If the ball strikes the bottom of the elevator shaf in 3 s, determine the height of the elevator from the bottom of the shaft at the instant the ball is released.Also, find the velocity of the ball when it strikes the bottom of the shaft Kinematics: When the ball is released, its velocity will be the same as the elevator at the instant of release. Thus, va-6 ft/s. Also, t=3s, sa-0, s=-h, and ac =-32.2 ft/s, -k-0 + 6(3) + (-32.2)(32) h = 127 ft Ans. v6t (-322)(3) -90.6 ft/s=90.6 ft/s

Explanation / Answer

here,

the initial velocity of ball is same as that of ball when it is released , v0 = 6 ft/s

time taken by ball , t = 3 s

s = - h

and accelration due to gravity , g = 9.8 m/s

g = 32.2 ft/s^2

using seccond equation of motion

s = v0*t + 0.5 * a * t^2

- h = 6 * 3 + 0.5 * (- 32.2) * 3^2

h = 127 ft

v = v0 + (- a*t)

v = 6 - 32.2 * 3

v = 90.6 feet/s downwards

here, constant accelration is due to gravity ,

i.e g = 9.8 m/s^2 = 32.2 m/s^2

as their is no other force acting on the ball

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

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