Answer:
a) [tex]v(2\,s) = 5.9\,\frac{m}{s}[/tex], [tex]v(4\,s) = -13.7\,\frac{m}{s}[/tex], b) [tex]t = 2.602\,s[/tex], c) [tex]h(2.602\,s) = 35.176\,m[/tex], d) [tex]t = 5.281\,s[/tex], e) [tex]v(2.602\,s) = -26.254\,\frac{m}{s}[/tex]
Explanation:
a) The velocity function is determined by deriving the position function in time:
[tex]v(t) = 25.5-9.8\cdot t[/tex]
Velocities after 2 seconds and 4 seconds are, respectively:
[tex]v(2\,s) = 5.9\,\frac{m}{s}[/tex]
[tex]v(4\,s) = -13.7\,\frac{m}{s}[/tex]
b) The maximum height is reached when velocity is equal to zero:
[tex]25.5-9.8\cdot t = 0[/tex]
The time when the projectile reaches the maximum height:
[tex]t = 2.602\,s[/tex]
c) The maximum height is:
[tex]h (2.602\,s) = 2 + 25.5\cdot (2.602\,s)-4.9\cdot (2.602\,s)^{2}[/tex]
[tex]h(2.602\,s) = 35.176\,m[/tex]
d) The projectile hits the ground when height is equal to zero:
[tex]-4.9\cdot t^{2}+25.5\cdot t + 2 =0[/tex]
The roots of the second order polynomial are presented below:
[tex]t_{1} \approx 5.281\,s[/tex]
[tex]t_{2} \approx -0.077\,s[/tex]
The first one is the only reasonable solution in physical terms.
[tex]t = 5.281\,s[/tex]
e) The velocity of the projectile when it hits the ground is:
[tex]v(2.602\,s) = 25.5-9.8\cdot (5.281\,s)[/tex]
[tex]v(2.602\,s) = -26.254\,\frac{m}{s}[/tex]
