1 When is the projectile 4000ft and 6500 ft above the ground

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?

Solution

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)t^2-1 + 72.00(1) + 0             since d/dx xⁿ = n x^n-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