A football game costumarily begins with a coin toss to determine who kicks off. The referee tosses the coin vertically upwards with an initial velocity of 5m/s and it takes 1.02s for it to be in the air. In the absence of air resistance
1.How high does the coin go above its point of release?
2.How long does it the coin to reach its highest point?
3. How long after being thrown will the velocity of the coin be 2m/s upwards?
4. When will the displacement of the coin be zero?
5. When is the magnitude of the coins velocity equal to half its initial velocity?
6. When is the magnitude of the coins displacement equal to half the greatest height to which it rises?
1. 0=5^2+2(-9.81)y
y=1.27m
2. 0=5+(-9.81)t
t=0.51s
3. 2=5+(-9.81)t
t=0.31s
4. 0=1.27+0t+1/2(-9.81)t^2
t^2=0.258
t=0.51s
The displacement =0 between 0.51s and 1.02s
5. 1/2(5)=5+(-9.81)t
t=0.25s
between 0.25s and 0.77s
6. y=1/2vt
1/2(1.27)=1/2(5)t
t=0.25s
between 0.25s and 0.77s
1. The coin goes 2.55 meters above its point of release. It's probably waving "hi" to the spectators up there.
2. The coin takes 0.51 seconds to reach its highest point. It's so quick, it could become a coin superhero!
3. The velocity of the coin will be 2m/s upwards approximately 1.27 seconds after being thrown. It's understandable if the coin feels a little down at that point.
4. The displacement of the coin will be zero when it reaches the ground. In other words, it hits the ground at the end of its triumphant flight. The game begins!
5. The magnitude of the coin's velocity is equal to half its initial velocity approximately 0.77 seconds after being thrown. It's like the coin is just caught up in the hype and slows down a bit to enjoy the moment.
6. The magnitude of the coin's displacement is equal to half the greatest height it reaches approximately 0.77 seconds after being thrown. It's like the coin wants to experience the thrill of being halfway up and halfway down at the same time. How adventurous!
To solve these problems, we can use the equations of motion for an object in free fall:
1. How high does the coin go above its point of release?
The height reached by the coin can be determined using the equation for displacement:
s = ut + (1/2)at^2
where
s = displacement (height above the point of release)
u = initial velocity (5 m/s)
t = time in air (1.02 s)
a = acceleration (in the absence of air resistance, a = -9.8 m/s^2)
Plugging in the values:
s = (5)(1.02) + (1/2)(-9.8)(1.02)^2
s = 5.1 - 5.00198
s ≈ 0.098 m
So, the coin goes approximately 0.098 meters or 9.8 cm above its point of release.
2. How long does it take the coin to reach its highest point?
The time taken to reach the highest point can be determined using the equation for vertical velocity:
v = u + at
where
v = final velocity (at the highest point, v = 0 m/s)
u = initial velocity (5 m/s)
a = acceleration (in the absence of air resistance, a = -9.8 m/s^2)
Plugging in the values:
0 = 5 - 9.8t
9.8t = 5
t = 5/9.8
t ≈ 0.51 s
So, it takes approximately 0.51 seconds for the coin to reach its highest point.
3. How long after being thrown will the velocity of the coin be 2 m/s upwards?
The equation to determine the time for a certain velocity can be given as:
v = u + at
where
v = final velocity (2 m/s)
u = initial velocity (5 m/s)
a = acceleration (in the absence of air resistance, a = -9.8 m/s^2)
Plugging in the values:
2 = 5 - 9.8t
9.8t = 5 - 2
9.8t = 3
t = 3/9.8
t ≈ 0.31 s
So, the velocity of the coin will be 2 m/s upwards approximately 0.31 seconds after it is thrown.
4. When will the displacement of the coin be zero?
The displacement of the coin will be zero when it is back at its point of release. To find the time, we can use the equation for displacement:
s = ut + (1/2)at^2
where
s = displacement (0 m at the point of release)
u = initial velocity (5 m/s)
t = time
a = acceleration (in the absence of air resistance, a = -9.8 m/s^2)
Plugging in the values:
0 = 5t + (1/2)(-9.8)t^2
0 = 5t - 4.9t^2
0 = t(5 - 4.9t)
t = 0 or t = 5/4.9
Since time cannot be negative, the displacement of the coin will be zero approximately at t = 5/4.9 s or 1.02 s.
5. When is the magnitude of the coin's velocity equal to half its initial velocity?
The equation to determine the time for a certain velocity can be given as:
v = u + at
where
v = final velocity (0.5 * 5 = 2.5 m/s)
u = initial velocity (5 m/s)
a = acceleration (in the absence of air resistance, a = -9.8 m/s^2)
Plugging in the values:
2.5 = 5 - 9.8t
9.8t = 5 - 2.5
9.8t = 2.5
t = 2.5/9.8
t ≈ 0.26 s
So, the magnitude of the coin's velocity will be half its initial velocity approximately 0.26 seconds after it is thrown.
6. When is the magnitude of the coin's displacement equal to half the greatest height to which it rises?
The magnitude of the displacement can be determined using the equation for displacement:
s = ut + (1/2)at^2
where
s = displacement (height above the point of release)
u = initial velocity (5 m/s)
t = time
a = acceleration (in the absence of air resistance, a = -9.8 m/s^2)
Half the greatest height can be calculated by dividing the answer from question 1 (0.098 m) by 2:
(1/2)(0.098) = 0.049 m
Plugging in the values:
0.049 = 5t + (1/2)(-9.8)t^2
0.049 = 5t - 4.9t^2
4.9t^2 - 5t + 0.049 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
where a = 4.9, b = -5, and c = 0.049
t = (-(-5) ± √((-5)^2 - 4(4.9)(0.049))) / (2(4.9))
t = (5 ± √(25 - 0.9604)) / (9.8)
t = (5 ± √24.0396) / 9.8
The positive root will give us the time after the coin is thrown:
t = (5 + √24.0396) / 9.8
t ≈ 1.396 s
So, the magnitude of the coin's displacement will be half the greatest height approximately 1.396 seconds after it is thrown.
To solve these problems, we can use the kinematic equations of motion associated with vertical motion. Let's break down each question and solve them one by one.
1. How high does the coin go above its point of release?
To find the maximum height, we need to use the equation:
v^2 = u^2 - 2as,
where v is the final velocity (0 m/s at the highest point), u is the initial velocity (5 m/s), a is the acceleration (approximately -9.8 m/s^2, considering gravity is acting downwards), and s is the displacement (height).
Rearranging the equation, we get:
s = (v^2 - u^2) / (2a)
s = (0^2 - 5^2) / (2 * -9.8)
s = (-25) / (-19.6)
s ≈ 1.28 meters
Therefore, the coin goes approximately 1.28 meters above its point of release.
2. How long does it take for the coin to reach its highest point?
To find the time taken to reach the highest point, we can use the equation:
v = u + at,
where v is the final velocity (0 m/s at the highest point), u is the initial velocity (5 m/s), a is the acceleration (approximately -9.8 m/s^2), and t is the time taken.
Rearranging the equation, we get:
t = (v - u) / a
t = (0 - 5) / (-9.8)
t ≈ 0.51 seconds
Therefore, it takes approximately 0.51 seconds for the coin to reach its highest point.
3. How long after being thrown will the velocity of the coin be 2 m/s upwards?
To find the time taken, we can use the equation:
v = u + at,
where v is the final velocity (2 m/s upwards), u is the initial velocity (5 m/s), a is the acceleration (approximately -9.8 m/s^2), and t is the time taken.
Rearranging the equation, we get:
t = (v - u) / a
t = (2 - 5) / (-9.8)
t ≈ 0.31 seconds
Therefore, the velocity of the coin will be 2 m/s upwards approximately 0.31 seconds after being thrown.
4. When will the displacement of the coin be zero?
The displacement of the coin will be zero when it reaches the same height from where it was thrown. This will occur at two points in the motion: when the coin is thrown (initially) and when it returns to the same height during its descent (after reaching the maximum height).
5. When is the magnitude of the coin's velocity equal to half its initial velocity?
To find the time, we can use the equation:
v = u + at,
where v is the final velocity (0 m/s at the highest point), u is the initial velocity (5 m/s), a is the acceleration (approximately -9.8 m/s^2), and t is the time taken.
Rearranging the equation, we get:
t = (v - u) / a
t = (0.5u - u) / a
t ≈ -0.5u / a
Since time cannot be negative in this context, we disregard the negative sign.
Therefore, the magnitude of the coin's velocity will be equal to half its initial velocity approximately 0.5 seconds after being thrown.
6. When is the magnitude of the coin's displacement equal to half the greatest height it rises?
The magnitude of the coin's displacement will be equal to half the greatest height it rises at two points: when it reaches the maximum height and when it reaches the same height during its descent. So, the time when the magnitude of the coin's displacement is equal to half the greatest height will occur at two different times during the motion.