Someone at a fifth-floor window (20 m above the ground) hurls a ball downward at

Question

Someone at a fifth-floor window (20 m above the ground) hurls a ball downward at 40.0° at a speed of 14.0 m/s.
(a) How long does the ball take to reach the sidewalk?
(b) How far from the base of the building does the ball strike the sidewalk?
(c) What is the horizontal component of the ball's velocity right before it strikes the sidewalk?
(d)What is the vertical component of the ball's velocity right before it strikes the sidewalk?
(e) How fast will it be traveling right before it strikes the sidewalk?

Explanation / Answer

v=14 m/s==>=40o==>vox=v*cos()10.725 m/s==>voy=v*sin()8.999 m/s

t=(-9+(92+1/2*4*9.8*20))/9.81.3 sec

a) 1.3 sec

b) vx*t=10.725*1.313.95 m=range

c) vox=vx=10.725 m/s or 10.7 m/s

d) -vyfinal=voy+g*t=9+9.8*1.3-21.75 m/s

e) (21.752+10.7252)24.25 m/s=vfinal

f) =tan(-yfinal/vx)-67.52o theta on the ground

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