Question

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?

Answer 1

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)=$(-16)t^{2} +24t+6$

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)=$(-16)t^{2} +24t+6$

h(0)=$(-16)(0)^{2} +24(0)+6$

h(0)=6 feet.

Therefore, Height of coin at time t=0 is 6.

For maximum height of the coin,

h(t)=$(-16)t^{2} +24t+6$

Differentiating both side,

$\frac{d}{dt}h(t)=\frac{d}{dt}[(-16)t^{2} +24t+6]$

$\frac{d}{dt}h(t)=\frac{d}{dt}[(-32)t+24]$

$\frac{d}{dt}h(t)=0$

$\frac{d}{dt}[(-32)t+24]=0$

t=$\frac{24}{32}$

t=$\frac{3}{4}$

t=0.75

Now,

h(t)=$(-16)t^{2} +24t+6$

h(0.75)=$(-16)(0.75)^{2} +24(0.75)+6$

h(0.75)=15 feet.

The maximum height of coin jake tosses is 15 feet.