The height (in meters) of a projectile shot vertically upward from a point 4 m above ground level with an initial velocity of 21.5 m/s is
h = 4 + 21.5t − 4.9t²
after t seconds. (Round your answers to two decimal places.)
(a) Find the velocity after 2 s and after 4 s.
(b) When does the projectile reach its maximum height? What is its maximum height?
h = 4 + 21.5t − 4.9t²
after t seconds. (Round your answers to two decimal places.)
(a) Find the velocity after 2 s and after 4 s.
(b) When does the projectile reach its maximum height? What is its maximum height?
-
h(t) = -4.9t^2 + 21.5t + 4
h'(t) = -9.8t + 21.5
(a)
h'(2) = -9.8•2 + 21.5 = 1.9m/s
h'(4) = -9.8•4 + 21.5 = -17.7m/s
(b)
The projectile reaches its maximum height when h'(t) = 0.
-9.8t + 21.5 = 0
9.8t = 21.5
t = 21.5/9.8 ≈ 2.19s
h(21.5/9.8) = -4.9(21.5/9.8)^2 + 21.5(21.5/9.8) + 4 ≈ 27.58m
h'(t) = -9.8t + 21.5
(a)
h'(2) = -9.8•2 + 21.5 = 1.9m/s
h'(4) = -9.8•4 + 21.5 = -17.7m/s
(b)
The projectile reaches its maximum height when h'(t) = 0.
-9.8t + 21.5 = 0
9.8t = 21.5
t = 21.5/9.8 ≈ 2.19s
h(21.5/9.8) = -4.9(21.5/9.8)^2 + 21.5(21.5/9.8) + 4 ≈ 27.58m