A ball with a horizontal speed of 1.25 m/s rolls off a bench 1.00 m above the floor.
a) How long will it take the ball to hit the floor?
b)How far from a point on the floor directly below the edge of the bench will the ball land?
I have absolutely no idea how to do this problem. I have all these kinematic formulas written down, but no clue what they mean. I love physics but my physics teacher just sucks at being a teacher...he is really really, really flaky.
a) How long will it take the ball to hit the floor?
b)How far from a point on the floor directly below the edge of the bench will the ball land?
I have absolutely no idea how to do this problem. I have all these kinematic formulas written down, but no clue what they mean. I love physics but my physics teacher just sucks at being a teacher...he is really really, really flaky.
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They really should start stating proper initial conditions. Gravity is NOT constant. Grr. (Sorry, pet peeve).
a)
Anyway, if g = -9.8m/sec^2
The formula is
Displacement = 1/2 a * t^2 + (Vo * t)
(you probably have this somewhere...)
Displacement is a distance.
a = g
t = time
Since vertical displacement and it rolled off horizontally, Vo = 0
==>
1 = 1/2 9.8 * t^2
t = sqrt( 2/9.8 ) sec
b)
Take the time you just calculated (t), since it's time of flight. Since there's no drag
(again with the initial conditions...grr.)
er.. no drag, the horizontal velocity doesn't change in flight. ==>
Distance = Velocity * Time.
or
Horizontal Distance = Horizontal Velocity * Time = 1.25m/s * t
a)
Anyway, if g = -9.8m/sec^2
The formula is
Displacement = 1/2 a * t^2 + (Vo * t)
(you probably have this somewhere...)
Displacement is a distance.
a = g
t = time
Since vertical displacement and it rolled off horizontally, Vo = 0
==>
1 = 1/2 9.8 * t^2
t = sqrt( 2/9.8 ) sec
b)
Take the time you just calculated (t), since it's time of flight. Since there's no drag
(again with the initial conditions...grr.)
er.. no drag, the horizontal velocity doesn't change in flight. ==>
Distance = Velocity * Time.
or
Horizontal Distance = Horizontal Velocity * Time = 1.25m/s * t
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Projectile motion consists of two motions.
One is a horizontal motion with constant velocity
The other is the vertical motion with constant downward acceleration equal to g .
The vertical height through which it falls is given as 1m,
Considering the vertical motion,
Its initial vertical speed is zero.
Acceleration = g = 9.8 m/s ²
Displacement vertically downward is 1m
Using the equation of motion,
The time to fall through a distance of 1 m is
H = 0 + ½ g t ²
1 = 4.9 t ²
t = 0.45s
Hence the answer to (a) is 0.45 s
--------------------------------------…
In this time , it moves horizontally a distance equal to horizontal velocity* time ( no acceleration )
S = ut = 1.25*0.45 = 0.5625 = 0.6 m
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One is a horizontal motion with constant velocity
The other is the vertical motion with constant downward acceleration equal to g .
The vertical height through which it falls is given as 1m,
Considering the vertical motion,
Its initial vertical speed is zero.
Acceleration = g = 9.8 m/s ²
Displacement vertically downward is 1m
Using the equation of motion,
The time to fall through a distance of 1 m is
H = 0 + ½ g t ²
1 = 4.9 t ²
t = 0.45s
Hence the answer to (a) is 0.45 s
--------------------------------------…
In this time , it moves horizontally a distance equal to horizontal velocity* time ( no acceleration )
S = ut = 1.25*0.45 = 0.5625 = 0.6 m
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myphysicsbuddy