M Wiley PLUS C Chegg C edugen.wileyplus.com/edugen/student/mainfr.uni Apps Hp Co

Question

M Wiley PLUS C Chegg C edugen.wileyplus.com/edugen/student/mainfr.uni Apps Hp Connected watch NHL Live Stre... Al watch NBA Live Str... ment FULL SCREEN PRINTER VERSION BACK NEXT Question 14 A plane, diving with constant speed at an angle of 48.9 with the vertical, releases a projectile at an altitude of 608 m. The projectile hits the ground 6.92 s after release. (a) What is the speed of the plane? (b) How far does the projectile travel horizontally during its flight? What were the magnitudes of the (c) horizontal and (d) vertical components of its velocity just before striking the ground? (State your answers to (c) and (d) as positive numbers.) (a) Number Units (b) Number Units (c) Number Units (d) Number Units SHOW HINT Question Attempts: 0 of 8 used SAVE FOR LATER SUBMIT ANSWER Copyright 2000-2016 by John Wiley & Sons, Inc. or related companies. All rights reserved. 2:55 PM 2/13/2016

Explanation / Answer

Suppose the speed of the plane at an altitude ( h = 608 m ) = u

u(x) = u sin48.9° and u(y) = u cos48.9°

(a) Using the following equation,

s = u*t + (1/2)*a*t^2

=> h = u(y)*t - (1/2)*g*t^2

=> 608 = (u cos48.9°)*6.92 - (1/2)*9.8*6.92^2 => u = 185.26 m/s

(b) The hrizontal distance travelled,

X = u(x)*t = (u sin48.9°)*6.92 = (185.26 sin48.9°)*6.92 = 966 m

(c) Since there is no acceleration in x-direction, so x-directional velocity will remain constant,

=>   v(x) = u(x) = u sin48.9° = 185.26*sin48.9° = 139.61 m/s

(d) Using the following equation with relevant signs,

v = u + a*t => -v(y) = -u(y) - g*t => v(y) = u(y) + g*t

=> v(y) = u cos48.9° + g*t = 185.26*cos48.9° + 9.8*6.92 = 189.6 m/s

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---

---

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