A ball is thrown upward with an initial velocity of 18.5 meters per second from an initial height of 22.5 meters.(assume acceleration is -9.8 m/s (1.find the position function giving the height us as a function of time T)(2. when does the ball hit the ground?)
the ball hits the ground when h=0
as before,
h(t) = -4.9t^2 + 18.5t + 22.5
so solve as usual.
To find the position function, we start with the equation of motion for an object under constant acceleration:
h(t) = h₀ + v₀t + (1/2)at²
Where:
- h(t) is the height of the ball at time t
- h₀ is the initial height (22.5 meters)
- v₀ is the initial velocity (18.5 m/s)
- a is the acceleration (-9.8 m/s²)
- t is the time in seconds
(1) Position function giving the height as a function of time:
h(t) = 22.5 + 18.5t + (1/2)(-9.8)t²
= 22.5 + 18.5t - 4.9t²
So, the position function giving the height of the ball as a function of time is h(t) = 22.5 + 18.5t - 4.9t².
(2) To find when the ball hits the ground, we need to determine the time it takes for the height to be zero. We can set h(t) = 0 and solve for t.
0 = 22.5 + 18.5t - 4.9t²
Rearranging the equation to the standard quadratic form:
4.9t² - 18.5t - 22.5 = 0
We can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b² - 4ac)) / 2a
Substituting the values:
t = (-(-18.5) ± √((-18.5)² - 4(4.9)(-22.5))) / (2 * 4.9)
Simplifying:
t = (18.5 ± √(342.25 + 441.0)) / 9.8
t = (18.5 ± √783.25) / 9.8
Taking the positive root:
t = (18.5 + √783.25) / 9.8
Calculating the value:
t ≈ 3.75 seconds
Therefore, the ball hits the ground approximately 3.75 seconds after it was thrown.
To find the position function for the height of the ball as a function of time (T), we can use the kinematic equation:
h(T) = h0 + v0 * T + (1/2) * a * T^2
Where:
h(T) is the height of the ball at time T
h0 is the initial height (22.5 meters)
v0 is the initial velocity (18.5 meters per second)
a is the acceleration (-9.8 m/s^2)
T is the time elapsed
1. Position function for height as a function of time:
Using the given values, the position function becomes:
h(T) = 22.5 + (18.5 * T) + (1/2) * (-9.8) * T^2
Simplifying, we have:
h(T) = 22.5 + 18.5T - 4.9T^2
So, the position function for the height of the ball as a function of time (T) is h(T) = 22.5 + 18.5T - 4.9T^2.
2. When does the ball hit the ground?
The ball hits the ground when the height becomes zero. We can set h(T) = 0:
0 = 22.5 + 18.5T - 4.9T^2
Rearranging the equation, we have:
4.9T^2 - 18.5T - 22.5 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or quadratic formula. In this case, I will use the quadratic formula:
T = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we get:
T = (18.5 ± √(18.5^2 - 4 * 4.9 * -22.5)) / (2 * 4.9)
Calculating the expression within the square root:
T = (18.5 ± √(342.25 + 441.0)) / 9.8
T = (18.5 ± √783.25) / 9.8
T = (18.5 ± 27.98) / 9.8
There are two possible solutions, one positive and one negative. Since time cannot be negative in this context, we discard the negative solution.
T = (18.5 + 27.98) / 9.8
T ≈ 4.68 seconds
Therefore, the ball hits the ground approximately 4.68 seconds after it was thrown.