To find when the roller coaster reaches its maximum height, we need to find the maximum value of the function ℎ ( 𝑡 )

0.3 𝑡 3 − 5 𝑡 2 + 21 𝑡 h(t)=0.3t 3 −5t 2 +21t, where 𝑡 t is the time in seconds and ℎ ( 𝑡 ) h(t) is the height in feet.

Step 1: Find the derivative of ℎ ( 𝑡 ) h(t) The derivative ℎ ′ ( 𝑡 ) h ′ (t) gives the rate of change of height, which we can set equal to zero to find the critical points (where the coaster reaches maximum or minimum height).

ℎ ′ ( 𝑡 )

𝑑 𝑑 𝑡 ( 0.3 𝑡 3 − 5 𝑡 2 + 21 𝑡 )

0.9 𝑡 2 − 10 𝑡 + 21 h ′ (t)= dt d ​ (0.3t 3 −5t 2 +21t)=0.9t 2 −10t+21 Step 2: Set the derivative equal to zero and solve for 𝑡 t To find the critical points, set ℎ ′ ( 𝑡 )

0 h ′ (t)=0:

0.9 𝑡 2 − 10 𝑡 + 21

0 0.9t 2 −10t+21=0 We can solve this quadratic equation using the quadratic formula:

𝑡

− 𝑏 ± 𝑏 2 − 4 𝑎 𝑐 2 𝑎 t= 2a −b± b 2 −4ac ​

​

where 𝑎

0.9 a=0.9, 𝑏

− 10 b=−10, and 𝑐

21 c=21.

First, calculate the discriminant:

Δ

( − 10 ) 2 − 4 ( 0.9 ) ( 21 )

100 − 75.6

24.4 Δ=(−10) 2 −4(0.9)(21)=100−75.6=24.4 Now, plug in the values to the quadratic formula:

𝑡

− ( − 10 ) ± 24.4 2 ( 0.9 )

10 ± 24.4 1.8 t= 2(0.9) −(−10)± 24.4 ​

​

1.8 10± 24.4 ​

​

Approximate the square root of 24.4:

𝑡

10 ± 4.94 1.8 t= 1.8 10±4.94 ​

This gives two possible values for 𝑡 t:

𝑡 1

10 + 4.94 1.8

14.94 1.8 ≈ 8.3 t 1 ​

1.8 10+4.94 ​

1.8 14.94 ​ ≈8.3 𝑡 2

10 − 4.94 1.8

5.06 1.8 ≈ 2.81 t 2 ​

1.8 10−4.94 ​

1.8 5.06 ​ ≈2.81 Since the time interval we're considering is from 𝑡

0 t=0 to 𝑡

10 t=10 seconds, both values 𝑡 1 ≈ 8.3 t 1 ​ ≈8.3 and 𝑡 2 ≈ 2.81 t 2 ​ ≈2.81 are within this range.

Step 3: Check whether these points correspond to a maximum To determine whether these points are maxima or minima, we can either check the second derivative or evaluate the heights at these critical points and endpoints.

Second derivative test: The second derivative of ℎ ( 𝑡 ) h(t) is:

ℎ ′ ′ ( 𝑡 )

𝑑 𝑑 𝑡 ( 0.9 𝑡 2 − 10 𝑡 + 21 )

1.8 𝑡 − 10 h ′′ (t)= dt d ​ (0.9t 2 −10t+21)=1.8t−10 For 𝑡 2

2.81 t 2 ​ =2.81:

ℎ ′ ′ ( 2.81 )

1.8 ( 2.81 ) − 10

5.06 − 10

− 4.94 h ′′ (2.81)=1.8(2.81)−10=5.06−10=−4.94 Since ℎ ′ ′ ( 2.81 ) < 0 h ′′ (2.81)<0, this point is a maximum.

For 𝑡 1

8.3 t 1 ​ =8.3:

ℎ ′ ′ ( 8.3 )

1.8 ( 8.3 ) − 10

14.94 − 10

4.94 h ′′ (8.3)=1.8(8.3)−10=14.94−10=4.94 Since ℎ ′ ′ ( 8.3 )

0 h ′′ (8.3)>0, this point is a minimum.

Step 4: Find the maximum height The roller coaster reaches its maximum height at 𝑡 ≈ 2.81 t≈2.81 seconds. Now, plug this value back into the original height function to find the maximum height:

ℎ ( 2.81 )

0.3 ( 2.81 ) 3 − 5 ( 2.81 ) 2 + 21 ( 2.81 ) h(2.81)=0.3(2.81) 3 −5(2.81) 2 +21(2.81) Calculate each term:

ℎ ( 2.81 ) ≈ 0.3 ( 22.2 ) − 5 ( 7.9 ) + 21 ( 2.81 ) h(2.81)≈0.3(22.2)−5(7.9)+21(2.81) ℎ ( 2.81 ) ≈ 6.66 − 39.5 + 59.01 ≈ 26.17  feet h(2.81)≈6.66−39.5+59.01≈26.17 feet Conclusion: The roller coaster reaches its maximum height at approximately 𝑡

2.81 t=2.81 seconds, and the maximum height is about 26.17 feet.