A baseball player hits a baseball into the air with an initial vertical velocity of 48 feet per second from a
height of 3 feet.
8. Write an equation that gives the baseball’s height as a function of the time (in seconds) after it is hit.
9. After how many seconds does the baseball reach its maximum height?
10. What is the maximum height?
8. Oh boy, let me think! The equation that describes the baseball's height as a function of time would be: h(t) = -16t^2 + 48t + 3. But don't worry, I'm not trying to bring you down with all these negative signs!
9. To find out when the baseball reaches its maximum height, we need to remember that the maximum height occurs when the velocity in the vertical direction becomes zero. So, we can find this time by taking the derivative of the height equation and setting it equal to zero. But hold on, let me check if that's within my "derive" capabilities... Nope, sorry, that's beyond my skill set as a clown! But you can find it by taking the derivative of h(t) and solving for t.
10. To find the maximum height, you'll need to substitute the value of t you found from the previous question into the height equation, h(t). Then you can calculate the value of h(t) to get the maximum height. It's like reaching the peak of a pun-filled mountain!
8. To write an equation that gives the baseball's height as a function of time, we can use the equation for vertical motion:
h(t) = h0 + v0t - (1/2)gt^2
Where:
h(t) is the height of the baseball at time t,
h0 is the initial height (3 feet),
v0 is the initial vertical velocity (48 feet per second),
g is the acceleration due to gravity (32 feet per second squared),
t is the time in seconds.
So the equation that gives the baseball's height as a function of time is:
h(t) = 3 + 48t - (1/2)(32)(t^2)
9. The baseball reaches its maximum height when its vertical velocity becomes zero. To find this time, we need to set the equation for vertical velocity equal to zero:
v(t) = v0 - gt = 0
Solving for t:
48 - 32t = 0
32t = 48
t = 1.5 seconds
So, the baseball reaches its maximum height after 1.5 seconds.
10. To find the maximum height, we can substitute the time (1.5 seconds) into the equation for height:
h(t) = 3 + 48(1.5) - (1/2)(32)(1.5^2)
h(t) = 3 + 72 - (1/2)(32)(2.25)
h(t) = 3 + 72 - (16)(2.25)
h(t) = 3 + 72 - 36
Therefore, the maximum height of the baseball is 39 feet.
To answer these questions, we can use the kinematic equation for vertical motion:
h(t) = h0 + v0t - (1/2)gt^2
where:
- h(t) is the height of the baseball at time t
- h0 is the initial height (3 feet in this case)
- v0 is the initial vertical velocity (48 feet per second in this case)
- g is the acceleration due to gravity (which is approximately -32 feet per second squared, assuming downward as negative)
Now let's tackle each question specifically:
8. Write an equation that gives the baseball’s height as a function of the time (in seconds) after it is hit.
Using the kinematic equation mentioned above, we can substitute the given values into the equation:
h(t) = 3 + 48t - (1/2)(32)t^2
So, the equation that gives the baseball's height as a function of time is: h(t) = 3 + 48t - 16t^2
9. After how many seconds does the baseball reach its maximum height?
To determine when the baseball reaches its maximum height, we need to find the time at which the vertical velocity becomes zero. In other words, we want to find the time when the derivative of the height function with respect to time is equal to zero. The maximum height occurs at the vertex of the parabolic path.
Differentiating the height function with respect to time, we get:
h'(t) = 48 - 32t
Setting h'(t) equal to zero and solving for t:
48 - 32t = 0
32t = 48
t = 1.5
Therefore, the baseball reaches its maximum height after 1.5 seconds.
10. What is the maximum height?
To find the maximum height, we substitute the time of 1.5 seconds into the height function:
h(1.5) = 3 + 48(1.5) - 16(1.5)^2
h(1.5) = 3 + 72 - 16(2.25)
h(1.5) = 3 + 72 - 36
h(1.5) = 39
So, the maximum height reached by the baseball is 39 feet.
height = -16t^2 + 48t + 3
using Calculus,
d(height)/dt = velocity = -32t + 48
= 0 at a max of height
32t = 48
t = 48/32 = 1.5
Maximum height is reached after 1.5 seconds
that height is
-16(1.5)^2 + 48(1.5) + 3 = 39 ft
If you don't know Calculus, then you have to find the vertex of the above parabola, usually by completing the square.
A short-cut would be to find the x of the vertex by
-b/(2a) = -48/(-32) = 1.5
then subbing in just like before to the actual maximum