Jake tosses a coin up in the air and lets it fall to the ground. The equation that models the height (in feet) and time (in seconds) of the parabola is h(t)= -16t^2+ 24t+6. What is the height of the coin when Jake tosses it?
Accepted Solution
A:
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=0Height 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=0.75Now,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.