A student throws a ball upward from a height of 48ft, initially at 32ft per second. What is the maximum height of the ball?
your function for the height is
h = -16t^2 + 32t + 48
find the vertex of this quadratic function using whatever method you have learned.
Since you titled your post "Calculus",
find the derivative, set it equal to zero and solve for t.
Finally, plug that t value back into my function.
To find the maximum height of the ball, we need to use the formula for the height of an object thrown vertically:
h(t) = h0 + v0*t - 16t^2
Where:
h(t) = height of the object at time t
h0 = initial height
v0 = initial velocity (in this case, the upward velocity)
t = time
Given:
h0 = 48ft (initial height)
v0 = 32ft/s (initial upward velocity)
To find the maximum height, we need to find the time when the object reaches its highest point. At the highest point, the object's velocity becomes 0.
First, let's find the time it takes for the object to reach its highest point by setting the velocity equal to 0:
0 = v0 - 16t
Solving for t:
16t = v0
t = v0 / 16
t = 32ft/s / 16
t = 2s
Now that we have the time it takes for the object to reach its highest point, we can substitute it into the height equation to find the maximum height:
h(t) = h0 + v0*t - 16t^2
h(t) = 48ft + 32ft/s * 2s - 16(2s)^2
h(t) = 48ft + 64ft - 16(4s^2)
h(t) = 48ft + 64ft - 256ft
h(t) = -144ft
The maximum height of the ball is -144ft. However, the "-" sign indicates that the ball is below the initial height, so the maximum height is actually 144ft above the initial height. Therefore, the maximum height of the ball is 144ft.
To find the maximum height of the ball, we need to understand the motion of the ball and its trajectory.
The motion of the ball can be described by the equation of motion for vertical motion:
h(t) = h0 + v0t - (1/2)gt^2
Where:
- h(t) is the height of the ball at time t
- h0 is the initial height of the ball (48ft)
- v0 is the initial velocity of the ball (32ft per second)
- g is the acceleration due to gravity, which is approximately 32.2ft per second squared (on Earth)
At the maximum height, the velocity of the ball becomes zero because it momentarily stops before falling back down. Therefore, we can set v(t) = 0 and solve for t to find the time it takes for the ball to reach its maximum height. We can then substitute this time value into the equation to find the maximum height.
Let's calculate:
0 = 32 - gt
Solving for t:
t = 32/g
Now substitute this value back into the original equation:
h(t) = h0 + v0t - (1/2)gt^2
h(t) = 48 + 32(32/g) - (1/2)g(32/g)^2
h(t) = 48 + 32(32/g) - (1/2)g(32^2/g^2)
Simplifying:
h(t) = 48 + 32(32/g) - (1/2)(32^2/g)
Now, we can calculate the value of g:
g ≈ 32.2ft/s^2
Substituting this value:
h(t) = 48 + 32(32/32.2) - (1/2)(32^2/32.2)
Simplifying further:
h(t) = 48 + 32 - (1/2)(32^2/32.2)
Calculating the value:
h(t) ≈ 48 + 32 - 16.84
h(t) ≈ 63.16ft
Therefore, the maximum height of the ball is approximately 63.16 ft.