1. How far up does the coin go?
A girl standing on top of a 22m high building throws a coin upwards with a speed of 8.8 m/s.
How far up does the coin go?
b) How much time it takes the coin to reach the maximum height?
C) What is the coin speed as it hits the ground?
Break it down:
The height of building. We show it with “h = 22m”
A girl standing on top of a 22-m-high
Initial vector points +y axis. We choose the point that the girl throws the coin as our origin or initial instant.
building throws a coin upwards
with a speed of 8.8 m/s.
Vi = + 8.8m/s, along +y axis
How far up does the coin go?
What is the maximum height the coin can reach?
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2. How far up does the coin go?
A girl standing on top of a 22m high building throws a coin upwards with a speed of 8.8 m/s.
How far up does the coin go?
b) How much time it takes the coin to reach the maximum height?
C) What is the coin speed as it hits the ground?
Break it down:
Time it reaches the maximum height. Time interval the coin travels from throwing point to the maximum height.
b) How much time
it takes the coin to reach the maximum height?
c) What is the coin speed as it hits the ground?
Velocity vector when it hits the ground is along the – y axis. The magnitude of this vector is the speed at ground surface.
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3. How far up does the coin go?
A girl standing on top of a 22m high building throws a coin upwards with a speed of 8.8 m/s.
How far up does the coin go?
b) How much time it takes the coin to reach the maximum height?
C) What is the coin speed as it hits the ground?
+ y
Solution:
Final instant
Vi = + 8.8m/s
g = 9.8 m/s2 Vf = 0
y =? t =?
Draw a diagram Vi
Initial instant
+ x
h = 22m
Ground surface
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4. Upward or above chosen origin. Displacement is 3.95m along +y axis.
A girl standing on top of a 22m high building throws a coin upwards with a speed of 8.8 m/s.
How far up does the coin go?
b) How much time it takes the coin to reach the maximum height?
C) What is the coin speed as it hits the ground?
Solution:
a) How far up does the coin go?
We solve this part from the instant the girl throws the coin up (initial instant) to the instant the coin reaches the maximum height (final instant).
Use appropriate equation
Vf is the final speed at the maximum height which is zero.
At this point the coin stops and starts moving back downward.
Vf2 - Vi2 = 2gy
 0 – (8.8)2 = 2 [(-9.8) * y]
 y = [- (8.8)2] / (-2 * 9.8) = (-77.44) / (-19.6)
Upward or above chosen origin. Displacement is 3.95m along +y axis.
 y = +3.95m
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5. How far up does the coin go?
A girl standing on top of a 22m high building throws a coin upwards with a speed of 8.8 m/s.
How far up does the coin go?
b) How much time it takes the coin to reach the maximum height?
C) What is the coin speed as it hits the ground?
Solution:
b) How much time it takes the coin to reach the maximum height?
Use appropriate equation
For this part of motion the initial and final speeds are given. Acceleration is gravity acceleration and known, then the better equation to find time will be;
Vf = at + Vi
 Vf = [- g * t] + Vi
We have known values; Vf = 0 and Vi = + 8.8m/s
 0 = [-9.8 * t] + [+ 8.8]  t = (8.8)/(9.8)
 t = 0.9s
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6. How far up does the coin go?
A girl standing on top of a 22m high building throws a coin upwards with a speed of 8.8 m/s.
How far up does the coin go?
b) How much time it takes the coin to reach the maximum height?
C) What is the coin speed as it hits the ground?
+ y
Solution:
C) What is the coin speed as it hits the ground?
To answer part C we can solve it in two ways;
Initial instant
+ x
Vi = - 8.8m/s
First solution: keep the chosen origin.
g = 9.8 m/s2
y =22 m
h = 22m
This way we know that when the coin is reaching this point again its velocity is Vi = 8.8m/s but with negative sign, Vi = - 8.8m/s
Final instant
We draw a diagram:
Vf = ?
Ground surface
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7. How far up does the coin go?
A girl standing on top of a 22m high building throws a coin upwards with a speed of 8.8 m/s.
How far up does the coin go?
b) How much time it takes the coin to reach the maximum height?
C) What is the coin speed as it hits the ground?
Solution:
Use equation Vf2 – Vi2 = 2ay for linear motion or Vf 2 – Vi 2 = 2gh
Vf =?
Vf 2 – (+8.8)2 = 2[(-9.8) * (-22)]
Vf2 = =
Vf = 22.55m/s
Vf ~ 22.6m/s down or Vf ~ m/s
Vf = m/s
Magnitude
(Speed)
Magnitude + direction
(Velocity)
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8. How far up does the coin go?
A girl standing on top of a 22m high building throws a coin upwards with a speed of 8.8 m/s.
How far up does the coin go?
b) How much time it takes the coin to reach the maximum height?
C) What is the coin speed as it hits the ground?
+ y
Initial instant
Vi = 0
+ x
Solution:
C) What is the coin speed as it hits the ground?
Second solution: choose new origin.
g = 9.8 m/s2
y =25.95 m
h = 22m
Another way to solve this part is moving the origin to the max height. This way Vi = 0 which is velocity at max height, but the distance that coin travels equals to the sum of max height calculated in part “a” and the height of the building h = 22m then y = - 22m m or y = m (25.95m below new origin)
Vf = ?
Final instant
Ground surface
We draw a diagram:
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9. How far up does the coin go?
A girl standing on top of a 22m high building throws a coin upwards with a speed of 8.8 m/s.
How far up does the coin go?
b) How much time it takes the coin to reach the maximum height?
C) What is the coin speed as it hits the ground?
Solution:
C) What is the coin speed as it hits the ground?
Second solution:
Use appropriate equation
Vf 2 – Vi 2 = 2ay or Vf 2 – Vi 2 = 2gy
Vf 2 – 0 = 2(-9.8m/s2) * (-25.95m)
Vf 2 =
 Vf = 22.55m/s
Vf ~ 22.6m/s down or Vf ~ m/s
Magnitude + direction
(Velocity)
Magnitude
(Speed)
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