Answer:
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Step-by-step explanation:
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Answer:
<h3>A) 204m</h3><h3>B) 188m</h3>
Step-by-step explanation:
Given the rocket's height above the surface of the lake given by the function h(t) = -16t^2 + 96t + 60
The velocity of the rocket at its maximum height is zero
v = dh/dt = -32t + 96t
At the maximum height, v = 0
0 = -32t + 96t
32t = 96
t = 96/32
t = 3secs
Substitute t = 3 into the modeled function to get the maximum height
h(3) = -16(3)^2 + 96(3) + 60
h(3) = -16(9)+ 288 + 60
h(3) = -144+ 288 + 60
h(3) = 144 + 60
h(3) = 204
Hence the maximum height reached by the rocket is 204m
Get the height after 2 secs
h(t) = -16t^2 + 96t + 60
when t = 2
h(2) = -16(2)^2 + 96(2) + 60
h(2) = -64+ 192+ 60
h(2) = -4 + 192
h(2) = 188m
Hence the height of the rocket after 2 secs is 188m
Answer:
3(6+2) is the correct answer.
It gives function: h = -16t^2 + 40
water is when height is 0 so solve for 0= -16t^2 + 40
subtract 40 from both sides: -40 = -16t^2
divide by -16: -40/-16 = t^2
simplify: 5/2 = t^2
square root both sides: 1.58 = t
the orange hits the river in 1.58 seconds.
