A diver followed a path defined by h(t) = −4.9t 2 + 3t + 10in her dive, where t is the time,
in seconds, and h represents her height above the water, in metres.
a) What was the maximum height the diver reached?
b) For how long was the diver in the air?
Answers
Given : A diver followed a path defined by h(t) = −4.9t 2 + 3t + 10
in her dive, where t is the time,
in seconds, and h represents her height above the water, in metres.
To Find :a) What was the maximum height the diver reached?
b) For how long was the diver in the air?
Solution:
h(t) = - 4.9t² + 3t + 10
at t = 0 h = 10 m
Hence diver is 10 m above the surface of water
h'(t) = -9.8t + 3
=> t = 3/9.8
h''(t) = -9.8 < 0
Hence max height at t = 3/9.8
h(t) = - 4.9(3/9.8)² + 3(3/9.8) + 10
= - 9/19.6 + 9/9.8 + 10
= 10 + 9/19.6
= 10.459 m is max height
when h = 0 then diver is no more on air
Hence - 4.9t² + 3t + 10 = 0
=> t =( - 3 ± √(9 - 4(-4.9)(10)) / ( 2 * (-4.9)
only considering positive value
t = 1.767
Hence for 1.767 secs diver was in air
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