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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 solve these problems, we can start by using the equations of motion for an object in free fall:
1. Displacement (d) in terms of initial velocity (v₀), time (t), and acceleration due to gravity (g) is given by:
d = v₀t - 0.5gt²
2. The final velocity (v) in terms of initial velocity (v₀), time (t), and acceleration due to gravity (g) is given by:
v = v₀ - gt
Now let's solve each part of the problem step by step:
a.) To find how high the ball goes, we need to determine the maximum height reached by the ball. When the ball reaches its maximum height, its final velocity will be zero. Let's assume the upward direction as positive.
Using equation (2), we have:
v = v₀ - gt
At the maximum height, v = 0, so we can solve for t:
0 = 15.0 m/s - 9.8 m/s² * t
Rearranging the equation:
9.8 m/s² * t = 15.0 m/s
Dividing both sides by 9.8 m/s²:
t = 15.0 m/s / 9.8 m/s²
t ≈ 1.53 s
Now, we can find the height (d) using equation (1), substituting the time t=1.53s:
d = v₀t - 0.5gt²
d = 15.0 m/s * 1.53 s - 0.5 * 9.8 m/s² * (1.53 s)²
d ≈ 11.63 m
Therefore, the ball goes approximately 11.63 meters high.
b.) To find the time the ball is in the air before it comes back to the person's hand, we need to determine the total time it takes for the ball to reach its initial height again. Since the ball was thrown upward, it will take twice the time calculated in part (a) for the ball to return to the initial height.
t_total = 2 * t ≈ 2 * 1.53 s
t_total ≈ 3.06 s
Therefore, the ball is in the air for approximately 3.06 seconds before it comes back to the person's hand.
c.) To find the time(s) when the ball passes a point 8.00 m above the ground, we can use equation (1) and solve for t, but with d = 8.00 m:
8.00 m = v₀t - 0.5gt²
Rearranging the equation and setting the equation equal to zero:
0.5gt² - v₀t + 8.00 m = 0
This is a quadratic equation, which can be solved using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
In this case:
a = 0.5g
b = -v₀
c = 8.00 m
Substituting these values into the formula gives:
t = (-(-15.0) ± √((-15.0)² - 4 * 0.5 * 9.8 * 8.00)) / (2 * 0.5 * 9.8)
Simplifying:
t = (15.0 ± √(225.0 - 156.8)) / 9.8
t ≈ (15.0 ± √68.2) / 9.8
Calculating the two possible solutions:
t₁ ≈ (15.0 + √68.2) / 9.8 ≈ 2.35 s
t₂ ≈ (15.0 - √68.2) / 9.8 ≈ -0.94 s
Both solutions are valid because they represent different points in the ball's trajectory. t₁ is the time when the ball is going up and crosses the 8.00 meter mark, while t₂ is the time when the ball is coming down and crosses the same point.