a) The equation is s = 64t - 16.1 t² + 3
b) The ball will take 1.99 seconds to reach its maximum height
c) The maximum height the ball will reach is 66.60 feet
d) The ball will be in the air for 4.02 seconds
Step-by-step explanation:
* Lets revise some rules of distance , velocity and time
- If the displacement is s , initial velocity is u , time is t and acceleration of
free fall is a , then the equation of the trajectory is
s = ut - 1/2 at² ⇒ upward thrown
∵ The acceleration of free fall a = 32.2 feet/sec²
∴ s = ut - 1/2 (32.2) t²
∴ s = ut - 16.1 t²
∵ The initial velocity is 64 feet/second
∴ s = 64t - 16.1 t²
∵ The ball is thrown from a height 3 feet
∴ s - 3 = 64t - 16.1 t² ⇒ add 3 to both sides
∴ s = 64t - 16.1 t² + 3
a) The equation is s = 64t - 16.1 t² + 3
- The ball will reach the maximum height when its velocity (v)
reached to 0
∵ v = ds/dt
∵ s = 64t - 16.1 t² + 3
- The rule of differentiation
# y = ax^n ⇒ dy/dx = a(n) x^(n-1)
# y = ax ⇒ dy/dx = a
# y = a ⇒ dy/dx = 0
∴ ds/dt = 64 - 16.1(2) t + 0
∴ v = 64 - 32.2 t
∵ At maximum height v = 0
∴ 0 = 64 - 32.2 t ⇒ add 32.2 t to both sides
∴ 32.2 t = 64 ⇒ divide both sides by 32.2
∴ t = 1.987577 ≅ 1.99 seconds
b) The ball will take 1.99 seconds to reach its maximum height
- To find the maximum height substitute this value of t in the
equation of trajectory
∵ s = 64t - 16.1 t² + 3
∴ s maximum = 64(1.99) - 16.1(1.99)² + 3 = 66.60
c) The maximum height the ball will reach is 66.60 feet
- To find the time that the ball in the air put s = 0, because the ball will
return to the point of thrown
∵ s = 0
∵ s = 64t - 16.1 t² + 3
∴ 0 = 64t - 16.1 t² + 3 ⇒ multiply both sides by -1
∴ 16.1 t² - 64t - 3 = 0
- Use your calculator to factorize it and find the value of t
∴ t = 4.02 or -0.5
- We will refused the negative answer because there is no negative time
∴ t = 4.02
d) The ball will be in the air for 4.02 seconds
