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A stone is thrown into the air. Its height above the ground t seconds after it is thrown is given by

the function h(t) = -5t² + 30t+2 metres.
How high is the stone above the ground at time t = 3 seconds?
b From what height above the ground was the stone released?
At what time is the stone 27 m above the ground?

User Mark Lavin
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1 Answer

4 votes

Answer:

Step-by-step explanation:

a) To find the height of the stone above the ground at time t = 3 seconds, we need to substitute t = 3 into the function h(t) = -5t² + 30t + 2 and solve for h(3).

h(3) = (-53²) + (303) + 2 = -45 + 90 + 2 = 37m

b) To find from what height above the ground the stone was released, we need to find the value of h(0), which represents the height of the stone at time t = 0, which is the time the stone was released.

h(0) = (-50²) + (300) + 2 = 2 m

c) To find the time at which the stone is 27 m above the ground, we need to solve for t in the equation h(t) = 27.

27 = (-5t²) + (30t) + 2

27 = -5t² + 30t + 2

25 = -5t² + 30t

5t² - 30t - 25 = 0

(5t-5)(t+5) = 0

t = 1 or t = -5

As time cannot be negative, the time at which the stone is 27 m above the ground is t = 1 sec.

User NemPlayer
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