Aplane, diving downward at an angle of 60 degrees below the horizontal, releases

Question

Aplane, diving downward at an angle of 60 degrees below the horizontal, releases a sugar sack at an altitude of 730m. The sugar sack hits the ground 5.0s after its release.
a) What was the horizontal component of the sugar sack's velocity just before striking the groud?
b) What was the vertical component of the sugar sack's velocity just before striking the ground?
c) What was the speed of the plane?
d) How far did the sugar sack travel horizontally during its flight?
(hint, you can round all anwers to the nearest whole number)

Explanation / Answer

a) first we need to figure out the vertical velocity
h=(1/2)gt2+vyt

730=(1/2)(9.8)(52)+vy(5)

vy=121.5 m/s (downward)

now we can use trig to figure out the x component

vx=vy/tan

vx=121.5/tan60

vx=70.148 m/s

b) solve in a

C)

v=(vx2+vy2)

v=140.296 m/s

(you could also use trig again)

d)now we simply take the product of the horizontal velocity and the time

x=vxt

x=70.148(5)

x=350.74m

hope that helps

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