Answer:
The maximum height of coin Jake tosses is 15 feet.
Step-by-step explanation:
Given Jake tosses a coin up in the air
and the height of coin is model by h(t)=[tex](-16)t^{2} +24t+6[/tex]
To find height of the coin when Jake tosses it:
When a coin is in the hand of Jake, time t=0
Height of coin is h(t)=[tex](-16)t^{2} +24t+6[/tex]
h(0)=[tex](-16)(0)^{2} +24(0)+6[/tex]
h(0)=6 feet.
Therefore, Height of coin at time t=0 is 6.
For maximum height of the coin,
h(t)=[tex](-16)t^{2} +24t+6[/tex]
Differentiating both side,
[tex]\frac{d}{dt}h(t)=\frac{d}{dt}[(-16)t^{2} +24t+6][/tex]
[tex]\frac{d}{dt}h(t)=\frac{d}{dt}[(-32)t+24][/tex]
[tex]\frac{d}{dt}h(t)=0[/tex]
[tex]\frac{d}{dt}[(-32)t+24]=0[/tex]
t=[tex]\frac{24}{32}[/tex]
t=[tex]\frac{3}{4}[/tex]
t=0.75
Now,
h(t)=[tex](-16)t^{2} +24t+6[/tex]
h(0.75)=[tex](-16)(0.75)^{2} +24(0.75)+6[/tex]
h(0.75)=15 feet.
The maximum height of coin jake tosses is 15 feet.
