Question

A ball is thrown in the air vertically from ground level with initial velocity 240 ft/s. Find the average height of the ball over the time interval extending from the time of the ball's release to its return to ground level. The height at time t is h(t)=240t−16t2. Your answer must include units .

Answer 1

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

Answer 2

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.

What is velocity of a object?

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,

$h(t)=240t-16t^2.$

As at the ground level the value of the height is equal to the zero. Thus the equation becomes,

$0=240t-16t^2.\\16t^2=240t\\16t=240\\t=15\rm s$

Integrate the given equation with respect to time to find the average height as,

$H_{avg}=\dfrac{1}{b-a}\int\limits^a_b {240t-16t^2} \, dt \\H_{avg}=\dfrac{1}{15-0}\int\limits^0_15 {240t-16t^2} \, dt \\H_{avg}=\dfrac{1}{15-0}\times (120\times15^2- 5.33\times15^3)\\H_{avg}=600.08$

Hence the average height of the ball is 600.08 ft.

Learn more about the velocity here;

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