1. When is the projectile 40.00ft and 65.00 ft above the ground and when does it
ID: 3424030 • Letter: 1
Question
1. When is the projectile 40.00ft and 65.00 ft above the ground and when does it achieve it's maximum height?
A projectile is thrown in the vertical direction from a window (40.00 ft above the ground) on the third floor of a building. The height of the projectile satisfies the quadratic function h (t)= -16.00t^2 + 72.00t + 40.00 where h(t) denotes the height (measured in ft) above the ground of the projectile t seconds after it is thrown. When is the projectile 40.00ft and 65.00 ft above the ground and when does it achieve it's maximum height?Explanation / Answer
h(t) = -16.00t2 + 72.00t + 40.00
a) 40.00ft ==> h(t) = 40.00
==> 40.00 = -16.00t2 + 72.00t + 40.00
==> -16.00t2 + 72.00t = 0
==> t(-16.00t + 72.00) = 0
==> t = 0 , t = 72.00/16.00
==> t = 0 , t = 4.5
Hence the projectile is 40 feet above ground at t = 0 seconds (i.e, in the beginning) and at t = 4.5 seconds.
b) 65.00ft ==> h(t) = 65.00
==> 65.00 = -16.00t2 + 72.00t + 40.00
==> 16.00t2 - 72.00t + 65.00 - 40.00 = 0
==> 16.00t2 - 72.00t + 25.00 = 0
==> (4.00t)2 - 2(4.00t)(9) + 92 - 92 + 25.00 = 0
==> (4.00t - 9)2 = 81 - 25.00
==> (4.00t - 9)2 = 56
==> 4.00t - 9 = 56
==> 4.00t - 9 = 7.48 , -7.48
==> 4.00t = 9+7.48 , 9 - 7.48
==> 4.00t = 16.48 , 1.52
==> t = 4.12 , 0.38
Hence parojectile is 65 ft above ground at t = 0.38 seconds and t = 4.12 seconds
c) At maximum height h '(t) = 0
h(t) = -16.00t2 + 72.00t + 40.00
==> h '(t) = -16.00(2)t2-1 + 72.00(1) + 0 since d/dx xn = n xn-1
==> h '(t) = -32.00t + 72.00
h '(t) = 0 ==> -32.00t + 72.00 = 0
==> t = 72.00/32.00
==> t = 2.25
Hence projects is achieves maximum height at t = 2.225 seconds
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