Answer:
a) [tex]v(0) = 79 \frac{ft}{s}[/tex], b) [tex]v_{max} = v(0) = 79 \frac{ft}{s}[/tex], c) [tex]x(2.469) = 145.516 ft[/tex], d) [tex]t = 2.469 s[/tex], e) [tex]t_{2} = 5.485 s[/tex], f) [tex]h_{max} = 145.516 ft[/tex]
Step-by-step explanation:
a) Initial velocity
[tex]v(0) = - 32 \cdot (0) + 79\\v(0) = 79 \frac{ft}{s}[/tex]
b) Maximum velocity
Due to the negative slope and positive intercept, it is concluded that object's maximum velocity is the initial velocity found in previous point.
[tex]v_{max} = v(0) = 79 \frac{ft}{s}[/tex]
c) Maximum displacement
First, the displacement function, measured in feets, has to be found by integrating the velocity function:
[tex]x(t) = - 16 \cdot t^{2} + 79 \cdot t + C_{1}[/tex]
The initial condition is [tex]x(0) = + 48 ft[/tex]. Then:
[tex]C_{1} = + 48[/tex]
The complete displacement function is presented as follows:
[tex]x(t) = -16\cdot t^{2} + 79 \cdot t + 48[/tex]
Besides, the acceleration function is determined by deriving the velocity function:
[tex]a(t) = -32[/tex]
This finding allows to infer the existence of a maximum displacement by applying Second Derivative Test. The critical point is found by equalizing the velocity function to zero and clearing time:
[tex]- 32 \cdot t +79 = 0\\t = 2.469 s[/tex]
Finally, the maximum displacement is obtained:
[tex]x(2.469) = 145.516 ft[/tex]
d) The maximum displacement occur at [tex]t = 2.469 s[/tex].
e) Let [tex]x(t)[/tex] equalize to zero and the two roots are found by using Second-Order Polynomial General Equation:
[tex]t_{1,2} = \frac{-79 \pm \sqrt{79^{2}-4\cdot(-16)\cdot (48)} }{2 \cdot (- 16)}[/tex]
Where [tex]t_{1} = - 0.546 s[/tex] and [tex]t_{2} = 5.485 s[/tex]. [tex]t_{2}[/tex] is the only solution that is physically reasonable.
f) Maximum height
The maximum height is given by the maximum displacement obtained in point d). Hence, [tex]h_{max} = 145.516 ft[/tex]
