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A person throws a ball upward into the air with an initial velocity of 15.0 m/s.
Calculate the following:
a.) how high it goes.
b.) how long the ball is in the air before it comes back to his hand.
c.) at what time(s) will the ball pass a point 8.00 m.
If you got multiples times, are both solutions valid? Why or why not?
To calculate the height the ball reaches, we can use the following equation of motion:
h = (V₀² - V²) / (2g)
Where:
h is the height
V₀ is the initial velocity
V is the final velocity (which is 0 when the ball reaches maximum height)
g is the acceleration due to gravity (approximately 9.8 m/s²)
Step 1: Identify the given information
V₀ = 15.0 m/s
g = 9.8 m/s²
Step 2: Substitute the values into the equation and solve
h = (15.0² - 0²) / (2 * 9.8)
h = (225 - 0) / 19.6
h = 11.5 m
Therefore, the ball reaches a height of 11.5 meters.
Now let's calculate how long the ball stays in the air before it comes back to the person's hand.
We can use the following equation:
t = (V - V₀) / g
Where:
t is the time
V₀ is the initial velocity
V is the final velocity (which is negative when the ball reaches the initial height again)
g is the acceleration due to gravity (approximately 9.8 m/s²)
Step 1: Identify the given information
V₀ = 15.0 m/s
V = -15.0 m/s (negative because it is in the opposite direction of the initial velocity)
g = 9.8 m/s²
Step 2: Substitute the values into the equation and solve
t = (-15.0 - 15.0) / 9.8
t = -30.0 / 9.8
t ≈ -3.06 s
Since time cannot be negative in this situation, we consider the absolute value of the time. Therefore, the time the ball is in the air before it comes back to the person's hand is approximately 3.06 seconds.
Lastly, let's find the time(s) at which the ball passes a point 8.00 meters.
We can use the following equation:
h = V₀t + (1/2)gt²
Where:
h is the height (which is 8.00 m)
V₀ is the initial velocity
t is the time
g is the acceleration due to gravity (approximately 9.8 m/s²)
Step 1: Identify the given information
h = 8.00 m
V₀ = 15.0 m/s
g = 9.8 m/s²
Step 2: Substitute the values into the equation and solve for t
8.00 = 15.0t + (1/2)(9.8)t²
Rearranging the equation:
(1/2)(9.8)t² + 15.0t - 8.00 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
Where:
a = (1/2)(9.8) = 4.9
b = 15.0
c = -8.00
Step 3: Substitute the values into the quadratic formula and solve for t
t = (-15.0 ± √(15.0² - 4(4.9)(-8.00))) / (2(4.9))
t = (-15.0 ± √(225 + 156.8)) / 9.8
t = (-15.0 ± √(381.8)) / 9.8
Using a calculator, we find two possible solutions for t:
t₁ ≈ -3.24 s
t₂ ≈ 0.19 s
Similar to the previous question, we discard the negative solution since time cannot be negative in this context. Therefore, the ball passes the point 8.00 m at approximately 0.19 seconds.
In summary:
a.) The ball reaches a height of 11.5 meters.
b.) The ball is in the air for approximately 3.06 seconds.
c.) The ball passes the point 8.00 m at approximately 0.19 seconds.