Asked by tamara
                A snowball rolls of a roof by accident. The roof is sloped 40.0
o
relative to the horizontal. The edge of
the roof is 14.0 m above the ground, and the speed of the snowball is 7.00 m/s as it rolls off the roof.
Ignore air resistance. Coincidentally, there is a group of management students standing 4.00 m from
the building.
(a) How far from the building does the snowball land?
(b) What is the speed of the snowball at impact?
(c) Did our snowball hit the management students, I mean were the management students ’accidently’
hit by the snowball?
            
            
        o
relative to the horizontal. The edge of
the roof is 14.0 m above the ground, and the speed of the snowball is 7.00 m/s as it rolls off the roof.
Ignore air resistance. Coincidentally, there is a group of management students standing 4.00 m from
the building.
(a) How far from the building does the snowball land?
(b) What is the speed of the snowball at impact?
(c) Did our snowball hit the management students, I mean were the management students ’accidently’
hit by the snowball?
Answers
                    Answered by
            tamara
            
    I got 6.91 meters for part a. Im not sure if that is right or not. Im not sure how to do part b and c.
    
                    Answered by
            Scott
            
    when it leaves the roof
vertical velocity ... -7.00 sin(40º)
horizontal velocity ... 7.00 cos(40º)
the horizontal velocity is constant
the vertical velocity is increased by gravity (acceleration)
h = -1/2 g t² + v₀ t + h₀ ... t is the time of flight
0 = -4.9 t² - 7.00 sin(40º) t + 14.0
... use the quadratic formula to find t
(a) 7.00 cos(40º) t
(b) use t to find the final vertical velocity
... -7.00 sin(40º) t - g t
... add the horizontal and vertical components of velocity
... use Pythagoras (a² + b² = c²)
(c) only if they are REALLY tall
    
vertical velocity ... -7.00 sin(40º)
horizontal velocity ... 7.00 cos(40º)
the horizontal velocity is constant
the vertical velocity is increased by gravity (acceleration)
h = -1/2 g t² + v₀ t + h₀ ... t is the time of flight
0 = -4.9 t² - 7.00 sin(40º) t + 14.0
... use the quadratic formula to find t
(a) 7.00 cos(40º) t
(b) use t to find the final vertical velocity
... -7.00 sin(40º) t - g t
... add the horizontal and vertical components of velocity
... use Pythagoras (a² + b² = c²)
(c) only if they are REALLY tall
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