iPad 3:53 PM 57% . edugen.wileyplus.com Courses-Blackboard Learn tudy & Practice

Question

iPad 3:53 PM 57% . edugen.wileyplus.com Courses-Blackboard Learn tudy & Practice Assignment Gradebook ORION PRENTER VERSION BACK Chapter 03, Problem 75 As preparation for this problem, review Conceptual Example 10. From the top of a cliff overlooking a lake, a person throws two stones, as shown in the drawing. The cliff is 30.5 m high. The two stones described have identical initial speeds of vo -14.9 m/s and are thrown at an angle -32.4, one below the horizontal and one above the horizontal, what is the distance between the points where the stones strike the water? Neglect air resistance oblem10t obiem 12 oblem 23 oblem 72 by Study Number Units the tolerance is +/-2% Click if you would like to Show Work for this question: Question Attempts: 0 of S used SAVE FOR LATER Copyright © 2000-2015 by John Wiley & Sons, tnc. or related companies. An rights reserved. Version 4.16.1.7

Explanation / Answer

when stone thrown upward direction

vertical velocity = 14.9sin(32.4) = 7.98 m/s

7.98^2 = 2gh

h = 3.25 m

3.25 = 1/2gt^2

3.25*2/9.8 = t^2

t = 0.81 s

The cliff is 30.5 m high, so 30.5 + 3.25

total height = 33.75 m

33.75 m = 1/2gt^2

33.75*2 /9.8 = t^2

t = 2.62 s

Total time in air = 2.62 + 0.81 = 3.43 s

14.9cos(32.4) = horizontal velocity = 12.58 m/s

d = vx *t

d = 12.58 * 3.43 = 43.15 m

Stone thrown down:

vy = 14.9sin(32.4 ) = 7.98 m/s

v^2 = u^2 + 2gh

v^2 = 7.98^2 + (2 x 9.8 x 22.2) =

v = 22.33 m/s

velocity^2 at bottom = 469.77, sq-rt = v = 21.67 m/s

30.5 = (7.98 + 22.33)/2 x t

t = 2.01 s

vx = 14.9cos(32.4) = horizontal velocity = 12.58 m/s

d = vx*t

d = 12.58 * 2.01

d = 25.29

distance = 43.15 - 25.29 = 17.86 m

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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