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

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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

time of flight for the stone1 which is thrown above horizontal is T1

let T2 be the time of flight for the stone2 which is thrown below the horizontal

then apply

-h = (u1*sin(32.4)*T)-(0.5*g*T1^2)

u1 = 14.9 m/s
and g = 9.8 m/s^2 and h = 30.5 m

4.9*T1^2-(14.9*sin(32.4))*T)-30.5 = 0

4.9*T1^2 - 7.98*T - 30.5 = 0

solving we get T1 = 3.43 S

and similarly

for stone2

-30.5 = (-14.9*sin(32.4)*T2) -(0.5*9.8*T2^2)

4.9*T2^2+(7.98*T2)-30.5 =0

solving we get

T2 = 1.81 S

then Horintal distance travelled by sdtone1 is R1 = u1*cos(32.4)*T1 = 14.9*cos(32.5)*3.43 = 43.2 m

by stone2 is R2 = 14.9*cos(32.4)*1.81 = 22.77 m

then required idstance is R1-R2 = 43.2-22.77 = 20.43 m

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