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Sagot :
Answer:
The maximum height of the golf ball h(t) = 156.25m
The maximum height of the golf ball at t = 3.125 seconds
Step-by-step explanation:
Step(i):-
Given
h(t)=-16t²+100t ...(i)
Differentiating equation (i) with respective to 't'
[tex]\frac{dh}{dt} = -16(2t) + 100 (1)[/tex]
-32t + 100 =0
⇒ -32 t = -100
[tex]t = \frac{100}{32} = \frac{25}{8} = 3.125[/tex]
t = 3.125 > 0
Step(ii):-
Given h(t) = -16t²+100t = 0
t ( -16 t + 100 ) =0
t =0 and -16t = -100
t =0 and t = 6.25
Step(iii):-
h(t) = -16t²+100t ...(i)
put t=0 in equation (i)
h(0) = 0
put t = 3.125 in equation (i)
h(t) = -16(3.125)²+100(3.125)
h (t) = 156.25
Put t = 6.25
h(t) = -16(6.25)²+100(6.25)
= 0
The maximum value h(t) = 156.25 at 3.125
final answer:-
The maximum height of the golf ball h(t) = 156.25m
The maximum height of the golf ball at t = 3.125 seconds
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