Advertisements
Advertisements
Question
A stone is thrown into air from the top of a building of height h m. The height of the stone (in metres) above the ground after t seconds is given by h (t) = 5t2 + 30t + 2; where 't' is the time from when the stone is thrown.

- How high will the stone be from the ground after 2 seconds?
- From what height, above the ground, the stone is thrown?
- At what time will the stone be 37 m above the ground?
Advertisements
Solution
(i) Given,
The height of the stone above the ground after t seconds: h (t) = 5t2 + 30t + 2
t = 2
Substituting value of t = 2 in h(t):
⇒ h(2) = 5(2)2 + 30(2) + 2
= 5(4) + 60 + 2
= 20 + 60 + 2
= 82 m.
Hence, the height after 2 seconds is 82 m.
(ii) The initial height of stone,
Substituting value of t = 0 in h(t):
⇒ h(0) = 5(0)2 + 30(0) + 2
= 2 m.
Hence, stone is thrown from a height of 2 m.
(iii) Given,
h(t) = 37
⇒ 5t2 + 30t + 2 = 37
⇒ 5t2 + 30t + 2 − 37 = 0
⇒ 5t2 + 30t − 35 = 0
⇒ 5(t2 + 6t − 7) = 0
⇒ t2 + 6t − 7 = 0
⇒ t2 + 7t − t − 7 = 0
⇒ t(t + 7) − 1(t + 7) = 0
⇒ (t − 1)(t + 7) = 0
⇒ (t − 1) = 0 or (t + 7) = 0
⇒ t = 1 or t = −7
Since time cannot be negative,
t = 1 second.
Hence, stone will be 37 m above the ground after 1 second.
