Answer:
the average height of the ball = 600.08ft
Explanation:
To find an average value of a function over a certain time interval, such value can be found using this general integral formula:
F = 1/(b-a) ∫_a^b▒f(x)dx
Step1: find the time t when H is 0ft
At ground level, t=0sec
At H=0ft,
0=240t – 16t2
Therefore, 16t2 = 240t (dividing both sides by t we have,)
t =240/16 = 15sec
Step 2: Find the average height using the formula below;
Havg. = 1/(b-a) ∫_a^b▒H(t)dt
Havg. = 1/(15-0) ∫_0^15▒(240t-16t^2 )dt
Havg. = 1/15 |〖240t〗^2/2-〖16t/3〗^3 |_0^15
Havg. = 1/15 [(120*〖15〗^2 )- (5.33*〖15〗^3)]
Havg. = 1/15×(27000-17998.88)
Havg. = 600.08ft
The velocity of a object is the change of distance or height with respect to time.
The average height of the ball is 600.08 ft.
The velocity of a object is the change of distance or height with respect to time.
Given information-
The initial velocity of the ball is 240 ft/s.
The height is given by the function of time as,
[tex]h(t)=240t-16t^2.[/tex]
As at the ground level the value of the height is equal to the zero. Thus the equation becomes,
[tex]0=240t-16t^2.\\16t^2=240t\\16t=240\\t=15\rm s[/tex]
Integrate the given equation with respect to time to find the average height as,
[tex]H_{avg}=\dfrac{1}{b-a}\int\limits^a_b {240t-16t^2} \, dt \\H_{avg}=\dfrac{1}{15-0}\int\limits^0_15 {240t-16t^2} \, dt \\H_{avg}=\dfrac{1}{15-0}\times (120\times15^2- 5.33\times15^3)\\H_{avg}=600.08[/tex]
Hence the average height of the ball is 600.08 ft.
Learn more about the velocity here;
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