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?
a. h = (Vf^3 - Vo^2) / 2g,
h = (0 - (15)^2) / -19.6 = 11.5m.
b. t(up) = (Vf - Vo) / g,
t(up) = (0 - 15) / -9.8 = 1.53s.
t(dn) = t(up) = 1.53s.
T = t(up) + t(dn) = 1.53 + 1.53 = 3.06s. = Time in flight.
c. d = Vot + gt^2 = 8m,
15t -4.9t^2 = 8,
-4.9t^2 + 15t -8 = 0,
Use Quadratic Formula to find t:
t = 0.69s.
a.) Well, let's see how high this ball can get. We can use a little physics to help us out here. When the ball reaches its highest point, its velocity will be zero. So, we can use that to find the time it takes to reach the top. Using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can set v to zero and solve for a. But hey, guess what? Clown Bot forgot to mention the acceleration is -9.8 m/s² due to gravity. Silly Clown Bot! Anyway, solving for time gives us t = u / a, which is 15.0 m/s / 9.8 m/s². Plug that into your calculator, and you'll find out how long it takes for the ball to reach its highest point.
b.) Now, to figure out how long the ball is in the air, we just need to double the time it took to reach the top. That's because it takes the same amount of time to go up as it does to come back down. Physics symmetry at its finest!
c.) To find out at what time the ball passes a point 8.00 m, we can use another fun equation: s = ut + 1/2 at², where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time. We know that s is 8.00 m, u is 15.0 m/s, and a is -9.8 m/s². Substitute those values and solve for t, and you'll find the time(s) when the ball passes that point. And yes, if you get multiple solutions, both are valid because physics doesn't discriminate - the ball can pass that point more than once. Just like a clown can keep making you laugh over and over again! Well, unless you're afraid of clowns, in which case I apologize for the traumatizing imagery.
To calculate the desired values, we need to use the equations of motion for motion under constant acceleration.
a.) To find how high the ball goes, we need to determine the maximum height reached during its trajectory. We can use the equation:
v^2 = u^2 + 2as
Where:
v = final velocity = 0 m/s (at maximum height)
u = initial velocity = 15.0 m/s
a = acceleration = -9.8 m/s^2 (negative due to the downward force of gravity)
s = displacement (height)
Substituting the known values, we can solve for s:
0 = (15.0)^2 + 2(-9.8)s
Simplifying the equation:
225 = -19.6s
Dividing both sides by -19.6:
s ≈ 11.5 m
Therefore, the ball reaches a height of approximately 11.5 meters.
b.) To calculate the time the ball is in the air before it comes back down, we can use the equation:
v = u + at
At the highest point, the velocity is 0 m/s. The initial velocity (u) is 15.0 m/s, and the acceleration (a) is -9.8 m/s^2. Solving for t:
0 = 15.0 + (-9.8)t
Rearranging the equation gives:
9.8t = 15.0
t ≈ 1.53 seconds
Therefore, the ball is in the air for approximately 1.53 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, we can use the equation of motion:
s = ut + (0.5)at^2
Rearranging the equation:
0.5at^2 + ut - s = 0
Substituting the given values:
0.5(-9.8)t^2 + 15.0t - 8.00 = 0
Solving this quadratic equation will give us the times at which the ball passes the point 8.00 m.
Using the quadratic formula:
t = (-15.0 ± √(15.0^2 - 4(0.5)(-9.8)(-8.00))) / (2(0.5)(-9.8))
Simplifying further:
t = (-15.0 ± √(225 + 78.4)) / (-9.8)
t = (-15.0 ± √(303.4)) / (-9.8)
Since √(303.4) ≈ 17.4, the equation becomes:
t = (-15.0 ± 17.4) / (-9.8)
Using the plus-minus symbol, we get two possible solutions:
1) t = (-15.0 + 17.4) / (-9.8) ≈ 0.24 seconds
2) t = (-15.0 - 17.4) / (-9.8) ≈ -3.37 seconds
Both solutions are valid, but -3.37 seconds is not meaningful in this context as it represents a time before the ball was thrown. Therefore, the valid solution is approximately 0.24 seconds.
In conclusion:
a.) The ball goes approximately 11.5 meters high.
b.) The ball is in the air for approximately 1.53 seconds before it comes back to the person's hand.
c.) The ball passes a point 8.00 meters at approximately 0.24 seconds.
To calculate the height the ball reaches, we can use the equation for vertical displacement.
a.) To find the maximum height it goes, we need to determine the time it takes to reach the highest point. We can use the equation:
vf = vi + at
Where:
vf = final velocity (which is zero at the highest point)
vi = initial velocity (given as 15.0 m/s)
a = acceleration (equal to -9.8 m/s^2, as gravity acts downwards)
t = time
Rearranging the equation to solve for time (t), we get:
t = (vf - vi) / a
Since vf = 0, the equation simplifies to:
t = -vi / a
Substituting the given values, we get:
t = -15.0 m/s / -9.8 m/s^2 ≈ 1.53 seconds
Now, to find the maximum height (h), we can use the equation:
h = vi * t + (1/2) * a * t^2
Substituting the values:
h = (15.0 m/s) * (1.53 s) + (1/2) * (-9.8 m/s^2) * (1.53 s)^2
≈ 11.47 meters
Therefore, the ball goes approximately 11.47 meters high.
b.) To determine the total time the ball is in the air before it comes back to the hand, we can consider the entire trajectory. The time it takes to reach the highest point is the same as the time it takes to return to the hand.
So, the total time the ball is in the air before it comes back to the hand is approximately 2 * 1.53 seconds ≈ 3.06 seconds.
c.) To find at what time(s) the ball passes a point 8.00 meters, we need to consider the ball's trajectory when going up and coming down separately.
When the ball is going up:
We can use the equation for vertical displacement:
h = vi * t + (1/2) * a * t^2
Rearranging the equation to solve for time (t), we get:
t = (-vi ± √(vi^2 - 2ah)) / a
Since we are looking for positive time, we use the minus sign (as it represents when the ball is going up). Substituting the values, we have:
t = (15.0 m/s - √(15.0 m/s)^2 - 2 * (-9.8 m/s^2) * 8.00 m) / -9.8 m/s^2
≈ 1.60 seconds
So, at around 1.60 seconds, the ball passes a point 8.00 meters while going up.
When the ball is coming down:
We use the same equation, but this time we consider the plus sign since the velocity is negative when coming down.
t = (-vi + √(vi^2 - 2ah)) / a
Substituting the values:
t = (15.0 m/s + √(15.0 m/s)^2 - 2 * (-9.8 m/s^2) * 8.00 m) / -9.8 m/s^2
≈ 4.13 seconds
So, at around 4.13 seconds, the ball passes a point 8.00 meters while coming down.
Both solutions are valid because during the ascent and descent, the ball follows the same path. However, keep in mind that these calculations assume no air resistance, and actual results may vary slightly due to real-world factors.