A rocket which has initial velocity of 64' per second and is on a 8' stand. The formula is H(t) = -16t²+64t +8. Find the height at 1 second. When does it reach its max height? . What is the max height? When does it hit the ground?

a) 56

b) 2 seconds

c) 72

d) 4.12seconds

Step-by-step explanation:

Given the height of the rocket modelled by the equation

H(t) = -16t²+64t +8 where

t is in seconds

a) To find the height at t= 1sec, we will simply substitute t = 1 into the equation as shown;

H(t) = -16t²+64t +8

H(1) = -16(1)²+64(1)+8

H(1) = -16+64+8

H(1) = 64-8

H(1) = 56

Hence the height after 1sec is 56

b) The velocity of the rocket at maximum height is zero, hence;

dH/dt = 0 (at maximum height)

dH/dt = -32t+64

0= -32t+64

32t =64

t =64/32

t =2secs

Hence the rocket reaches its maximum height after 2seconds

c) Substitute t =2 into the equation

H(2) =-16(2)²+64(2)+8

H(2) = -64+128+8

H(2) = 64+8

H(2) = 72

Hence the maximum height is 72

d) The rocket hits the ground when

h(t) =0

Substitute into the equation

0 = -16t²+64t+8

Divide through by -8

0 = 2t²-8t-1

From the equation a =2, b = -8 and c = -1

Using the general formula

t = -b±√b²-4ac/2a

t = -(-8)±√(-8)²-4(2)(-1)/2(2)

t = 8±√64+8/4

t =8±√72/4

t =8±8.49/4

t = 8+8.49/4 and 8-8.49/5

t =16.49/4 and -0.49/4

t = 4.12 and -0.12

Time cannot be negative

Hence t = 4.12secs

The rocket hits the ground 4.12secs later

no

step-by-step explanation: