im trying to help my son i have no ida about this stuff any help would be great When a ball is thrown up into the air, it makes the shape of a parabola. The equation S= -16t^2 + v*t + k gives the height of the ball at any time, t in seconds, where “v” is the initial velocity (speed) in ft/sec and “k” is the initial height in feet (as if you were on top of a tower or building).
Make up a scenario where a ball is thrown, shot, etc. into the air. You can choose any initial velocity (in feet/sec) and any initial height (in feet) of the ball, but include them in your written scenario. The ball can leave your hand, the top of a building, etc. so you can use many different values for the initial height.
Insert the chosen values for “v” and “k” into the formula listed above.
Use the formula to find the height of the ball at any two values of time, t, in seconds that you want. Show your calculations and put units on your final answer!
Provide a written summary of your results explaining them in the context of the original problem.
Please make sure that your answers make sense!
If your answer is negative, that means the ball already hit the ground, so choose a smaller value for time.
Think about a ball going up into the air, you might throw it or put in a cannon. If you throw a ball up into the air, it will not end up being 800 feet in the air if it leaves your hand at 5 feet. Therefore, you would need to adjust your initial velocity. You may want to research initial velocity (speed) to figure out what seems reasonable! (ex. Your 5 year old cannot throw a ball into the air with an initial velocity of 300 feet/sec) J
Scenario:
Let's imagine a scenario where a ball is thrown up into the air from the top of a building. The initial velocity (v) of the ball is 20 ft/sec, and the initial height (k) of the ball is 100 feet.
Using the formula S = -16t^2 + vt + k, we can calculate the height of the ball at different values of time, t.
1. Time = 1 second:
S = -16(1)^2 + 20(1) + 100
S = -16 + 20 + 100
S = 104 feet
2. Time = 2 seconds:
S = -16(2)^2 + 20(2) + 100
S = -64 + 40 + 100
S = 76 feet
In this scenario, at time t = 1 second, the ball is 104 feet above the initial height. At time t = 2 seconds, the ball is 76 feet above the initial height.
These calculations show that the ball reaches a maximum height and then starts to come down. The parabolic shape of the ball's path can be observed. Since the initial height is 100 feet, the ball eventually returns to the ground when S becomes negative.
Scenario: A ball is thrown up into the air from the ground level with an initial velocity of 30 ft/sec and an initial height of 10 feet.
To find the height of the ball at any time, we can use the formula S = -16t^2 + v*t + k, where S represents the height, t represents time, v represents the initial velocity, and k represents the initial height.
1. We can choose two different values for time, t, to find the height of the ball at those specific moments.
Let's calculate the height of the ball at t = 2 seconds and t = 4 seconds.
For t = 2 seconds:
S = -16(2)^2 + 30(2) + 10
S = -64 + 60 + 10
S = 6 feet
Therefore, at t = 2 seconds, the ball is at a height of 6 feet.
For t = 4 seconds:
S = -16(4)^2 + 30(4) + 10
S = -256 + 120 + 10
S = -126
Since the height is negative (-126 feet), it means the ball has already hit the ground.
2. In the context of the original problem, we can interpret the results as follows:
- At t = 2 seconds, the ball has reached a height of 6 feet above its initial position. This indicates that the ball is still ascending after being thrown up into the air.
- At t = 4 seconds, the calculated height is negative (-126 feet), which means the ball has already hit the ground. This implies that the ball took approximately 4 seconds to reach the ground after being thrown.
It's important to note that the initial velocity and initial height can greatly influence the ball's trajectory and behavior. Adjusting these values can result in different outcomes, so it's important to choose reasonable values that make sense in the given scenario.