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KatRina [158]
3 years ago
5

A ball is thrown upward and its height after t seconds can be described by formula f(t)=−10t^2+24t+5.6. Find the maximum height

the ball will reach.
Please answer!
Mathematics
2 answers:
Elza [17]3 years ago
8 0

Answer:

The maximum height reached by ball while throwing upward is 20 unit

Step-by-step explanation:

Given as :

The position of ball moving upward with time t sec is

f(t) = - 10 t² + 24 t + 5.6

Let the maximum height does the ball reach = H unit

<u>Now, For maximum condition</u>

\frac{\partial f(t)}{\partial t} = 0

So, \frac{\partial ( - 10 t² + 24 t + 5.6 )}{\partial t} = 0

Or, \frac{\partial (- 10 t^{2})}{\partial t} +  \frac{\partial 2 4 t }{\partial t} +  \frac{\partial 5.6}{\partial t} = 0

Or, ( - 10 ) × 2 t + 24 + 0 = 0

Or, - 20 t + 24 = 0

Or, 20 t = 24

∴   t = \dfrac{24}{20}

i.e t = \dfrac{6}{5}

Or, t = 1.2 sec

Again

<u>Double differentiation of the function </u>

i.e \frac{\partial^2 (-20t + 24)}{\partial t^2}

Or, -20 + 0

i.e - 20

∵ - 20 0

So, The maximum height is gained at t = 1.2 seconds

So, Maximum height  at ( t = 1.2 sec) = - 10 t² + 24 t + 5.6

Or, H = - 10 (1.2)² + 24 × 1.2 + 5.6.

Or, H = - 1.44 × 10 + 28.8 + 5.6

Or, H = - 14.4 + 28.8 + 5.6

∴   H = - 14.4 + 34.4

i.e H = 20 unit

So, The maximum height = H = 20 unit

Hence, The maximum height reached by ball while throwing upward is 20 unit . Answer

torisob [31]3 years ago
4 0

Answer:

<h2>The maximum height is 20 feet.</h2>

Step-by-step explanation:

The height of the ball after t seconds is given by f(t) = -10t^{2} + 24t +5.6.

For any function, at the maximum point, first order differentiate of the function will be 0, and the second order differentiate will be less than 0.

Now, \frac{d f(t)}{dt} = -20t +24.

As per the given condition, -20t +24 = 0\\t = \frac{24}{20} = \frac{6}{5}.

Whatever the value of t is, the second order differentiate of the function is -20, that is \frac{d^{2} f(t)}{dt^{2} } = -20 \leq 0.

Hence, at t = \frac{6}{5}, the height of the ball will be maximum.

f(\frac{6}{5} ) = -10(\frac{6}{5} )^{2} + 24\times\frac{6}{5} + 5.6 = 20

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