Respuesta :
Answer:
The maximum height is 20 feet.
Step-by-step explanation:
The height of the ball after t seconds is given by [tex]f(t) = -10t^{2} + 24t +5.6[/tex].
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, [tex]\frac{d f(t)}{dt} = -20t +24[/tex].
As per the given condition, [tex]-20t +24 = 0\\t = \frac{24}{20} = \frac{6}{5}[/tex].
Whatever the value of t is, the second order differentiate of the function is -20, that is [tex]\frac{d^{2} f(t)}{dt^{2} } = -20 \leq 0[/tex].
Hence, at t = [tex]\frac{6}{5}[/tex], the height of the ball will be maximum.
[tex]f(\frac{6}{5} ) = -10(\frac{6}{5} )^{2} + 24\times\frac{6}{5} + 5.6 = 20[/tex]
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
Now, For maximum condition
[tex]\frac{\partial f(t)}{\partial t}[/tex] = 0
So, [tex]\frac{\partial ( - 10 t² + 24 t + 5.6 )}{\partial t}[/tex] = 0
Or, [tex]\frac{\partial (- 10 t^{2})}{\partial t}[/tex] + [tex]\frac{\partial 2 4 t }{\partial t}[/tex] + [tex]\frac{\partial 5.6}{\partial t}[/tex] = 0
Or, ( - 10 ) × 2 t + 24 + 0 = 0
Or, - 20 t + 24 = 0
Or, 20 t = 24
∴ t = [tex]\dfrac{24}{20}[/tex]
i.e t = [tex]\dfrac{6}{5}[/tex]
Or, t = 1.2 sec
Again
Double differentiation of the function
i.e [tex]\frac{\partial^2 (-20t + 24)}{\partial t^2}[/tex]
Or, -20 + 0
i.e - 20
∵ - 20 [tex]<[/tex] 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
