Answer:
The maximum height of the cannonball 'h' = 256
The time will take for the cannonball to reach the ground 't' =4
Step-by-step explanation:
Explanation:-
The given equation h(t) = -16 t² + 128 t ...(i)
Differentiating equation(i) with respective to 't' we get
[tex]h^{l} (t) = \frac{dh}{dt} = -16 (2 t) +128 (1)[/tex]
[tex]h^{l} (t) = -16 (2 t) +128 (1) = 0[/tex]
[tex]-32 t +128 = 0[/tex]
- 32 t = -128
t = 4
Now
Again differentiating with respective to 'x'
[tex]h^{ll} (t) = \frac{d^{2} h}{dt^{2} } = -16 (2 ) < 0[/tex]
The function is Maximum at t = 4
The maximum value
h(t) =-16 t² + 128 t
h(4) = - 16 (4)² + 128(4) = 256
Conclusion:-
The maximum height of the cannonball 'h' = 256
The time will take for the cannonball to reach the ground 't' =4