3. A snowball rolls off a barn roof, that slope down at an angle of 40.0 o with

Question

3. A snowball rolls off a barn roof, that slope

down at an angle of 40.0o with an initial speed

of 7.00 m/s The edge of the roof is 14.0 m

above the ground Ignore air resistance to

answer the following questions. Show all

your work.

a) Calculate the horizontal and the vertical

components of the initial velocity.

b) How long does the snowball stay in the

air before it hits the ground below?

c) How far out from the edge of the barn does the

snowball strike the ground?  Is the (innocent)

bystander in danger?

d) Calculate the horizontal and the vertical

components of the velocity of the snowball

just before the moment of impact. Finally,

calculate the net speed and the

direction of its motion.

Explanation / Answer

a.)

Let Vx = the horizontal component of velocity V= 7m/s
Vx= V cos 40
Vx =7(.766)
Vx = 5.36 m/s

Let Vy = the vertical component of velocity V =7m/s
Vy= V sin 40
Vy= 7(.643)
Vy = 4.50 m/sec

b.) final vertical velocity(Vf) at edge

2aS = Vf^2 - Vi^2

Vi = Vy = the starting velocity ot the roof = 4.50 m/s
S = Y = 14m
a = g due to gravity = 9.8m/sec^2

2(-9.8)(-14) = Vf^2 - (-4.50)^2 {minus signs mean downward }
......274.4 = Vf^2 - (+19.8)
19.8 + 274.4 = Vf^2
Vf^2 = 274.4 + 19.8
= 294.2
Vf = 17.15 m/s  
at = Vf - Vi
(a)t = (- 17.15) - (-4.50)
at= - 17.15 + 4.50 = - 12.65

t = (- 12.65) / ( - 9.8)
t = 1.30 secs.

c.)To find the distance from the edge of the barn X
we can use the formula

X = (Vx)t
=(5.36)(1.30)

X = 6.97m from the edge of the barn.

d.)

Vx = 5.36 m/s

Vfy = 17.15 m/s  

net speed = sqrt ( Vx^2 + Vy^2)

= 17.968 m/s

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