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c Chegg.com M Wiley PLUS C edugen.wileyplus.com/edugen/student/mainfr.uni EE: Apps Un HP Connected Watch NHL Live Stre... Al watch NBA Live Str y & Practice Assignment Gradebook ORION Assignment BACK FULL SCREEN PRINTER VERSION Question 13 In the figure, you throw a ball toward a wall at speed 35.0 m/s and at angle eo 42.0 above the horizontal. The wall is distance d 15.0 m from the release point of the ball. (a) How far above the release point does the ball hit the wall? What are the (b) horizontal and (c) vertical components of its velocity as it hits the wall? (a) Number Unit (b) Number Unit (c) Number Unit SHOW HINT 12:40 PM 2/13/2016

Explanation / Answer

initial speed = u = 35 m/s at 42 degrees

using equations

v^2 = u^2 + 2*a*D

D = ut + 1/2*at^2

using horizontal component

ux = 35*cos42 = 26 m/s

vx = horizontal component of its velocity as it hits the wall

ax = 0

D = ux*t + 1/2*ax*t^2

15 = 26*t + 0

t = 15/26 sec

using equation

vx^2 = ux^2 + 2*ax*D

vx^2 = 26^2 + 0

vx = 26 m/s {Answer of b}

using vertical component

D = uy*t + 1/2*ay*t^2

uy = 35*sin42 = 23.42 m/s

ay = -g = - 9,8 m/s^2

D = 23.42*(15/26) + (1/2)*(- 9.8)*(15/26)^2 = 11.88 m {vertical height (Answer of a)}

using equation

vy^2 = uy^2 + 2*ay*D

vy^2 = 23.42^2 - 2*9.8*11.88

vy = 17.76 m/s {Answer of c}

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