Answer :
Answer:
a) [tex]s_{max} = 96\,ft[/tex], b) [tex]v(4.449\,s) = -78.368\,\frac{ft}{s}[/tex]
Explanation:
a) The maximum height is obtained with the help of the First and Second Derivative Tests:
First Derivative
[tex]v(t) = 64 - 32\cdot t[/tex]
[tex]64 - 32\cdot t = 0[/tex]
[tex]t = 2\,s[/tex]
Second Derivative
[tex]a(t) = -32[/tex] (absolute maximum)
The maximum height reached by the ball is:
[tex]s (2\,s) = 32 + 64\cdot (2\,s) - 16\cdot (2\,s)^{2}[/tex]
[tex]s_{max} = 96\,ft[/tex]
b) The time required by the ball to hit the ground is:
[tex]32+64\cdot t - 16\cdot t^{2} = 0[/tex]
[tex]-16\cdot (t^{2}-4\cdot t - 2) = 0[/tex]
[tex]t^{2}-4\cdot t - 2 = 0[/tex]
[tex](t -4.449)\cdot (t+0.449)\approx 0[/tex]
Just one root offers a solution that is physically reasonable:
[tex]t = 4.449\,s[/tex]
The velocity of the ball when it hits the ground is:
[tex]v(4.449\,s) = 64 - 32\cdot (4.449\,s)[/tex]
[tex]v(4.449\,s) = -78.368\,\frac{ft}{s}[/tex]
Answer:
a) 96ft
b) -78.4ft/s
Explanation:
Given the height of the ball modelled as s(t)=32+64t−16t²
a) at maximum height, the velocity of the ball is zero
Velocity = d(s(t))/dt = -32t+64
Since v = 0m/s at maximum height
0 = -32t+64
32t = 64
t = 64/32
t = 2seconds
This means it takes 2 seconds for the ball to reach its maximum height.
The maximum height reached will occur at t = 2secs
s(2) =32+64(2)-16(2)²
s(2) = 32+128-64
s(2) = 96ft
b) When the ball hits the ground the height = 0
s(t) = 0
-16t²+64t+32 = 0
t²-4t-2 = 0
Factorizing the expression:
-(-4)±√(-4)²-4(1)(-2)/2(1)
= 4±√16+8/2
= 4±√24/2
= (4+2√6)/2 and (4-2√6)/2
= 2+√6 and 2-√6
t= 4.45s and -0.45s
Substitute t = 4.45s into the equation for calculating the velocity
V(t) = -32t+64
V(4.45) = -32(4.45)+64
V(4.45) = -142.4+64
V(4.45) = -78.4ft/s