Answer:
a) [tex]v(2\,s) = 2.9\,\frac{m}{s}[/tex], b) [tex]v(4\,s) = -16.7\,\frac{m}{s}[/tex], c) [tex]t = 2.296\,s[/tex], d) [tex]h (2.296\,s) = 27.829\,m[/tex], e) [tex]t \approx 4.679\,s[/tex], f) [tex]v(4.679\,s) = -23.354\,\frac{m}{s}[/tex]
Step-by-step explanation:
The velocity function can be derived by the differentiating the height function:
[tex]v = 22.5-9.8\cdot t[/tex]
Velocities after 2 and 4 seconds are, respectively:
a) [tex]v(2\,s) = 2.9\,\frac{m}{s}[/tex]
b) [tex]v(4\,s) = -16.7\,\frac{m}{s}[/tex]
The maximum height is reached when velocity is zero. Then:
[tex]22.5-9.8\cdot t = 0[/tex]
c) [tex]t = 2.296\,s[/tex]
The maximum height is:
d) [tex]h (2.296\,s) = 27.829\,m[/tex]
The time required to hit the ground is:
[tex]-4.9\cdot t^{2}+22.5\cdot t +2 = 0[/tex]
Roots of the second-order polynomial are:
[tex]t_{1} \approx 4.679\,s[/tex]
[tex]t_{2} \approx -0.087\,s[/tex]
Only the first root is physically reasonable.
e) [tex]t \approx 4.679\,s[/tex]
The velocity when the projectile hits the ground is:
f) [tex]v(4.679\,s) = -23.354\,\frac{m}{s}[/tex]
