Answer:
a) 56
b) 2 seconds
c) 72
d) 4.12seconds
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