Answer:
Approximately [tex]0.21\; {\rm s}[/tex] and then approximately [tex]4.8\; {\rm s}[/tex], assuming that air resistance on the stone is negligible and that [tex]g = -10\; {\rm m\cdot s^{-2}}[/tex].
Explanation:
Let [tex]h[/tex] denote the height of the stone.
Let [tex]g[/tex] denote the gravitational acceleration.
Let [tex]v_{0}[/tex] denote the initial velocity of the stone. In this question, [tex]v_{0} = 25\; {\rm m\cdot s^{-1}}[/tex].
Let [tex]t[/tex] denote the time (measured in seconds) after the stone was thrown.
The following SUVAT equation relates these quantities:
[tex]\displaystyle h = \frac{1}{2}\, g\, t^{2} + v_{0}\, t[/tex].
The value of height [tex]h[/tex], gravitational acceleration [tex]g[/tex], and initial velocity [tex]v_{0}[/tex] are all known: [tex]h = 5\; {\rm m}[/tex], [tex]g = (-10\; {\rm m\cdot s^{-2}})[/tex], and [tex]v_{0} = 25\; {\rm m\cdot s^{-1}}[/tex]. Substitute these values into the equation to obtain a quadratic equation about time [tex]t[/tex] (in seconds):
[tex]\displaystyle 5 = \frac{(-10)}{2}\, t^{2} + 25\, t[/tex].
Simplify this equation to obtain:
[tex]t^{2} - 5\, t + 1 = 0[/tex].
Solve this equation using the quadratic formula. The two roots of this quadratic equation about [tex]t[/tex] are:
[tex]\displaystyle t_{1} \approx 0.20871 \; {\rm s}[/tex], and
[tex]\displaystyle t_{2} \approx 4.79129 \; {\rm s}[/tex].
Both roots are greater than [tex]0[/tex] and are valid.
- The object would be at a height of [tex]5\; {\rm m}[/tex] above the ground at [tex]\displaystyle t_{1} \approx 0.20871 \; {\rm s}[/tex] after launch on its way up.
- At [tex]\displaystyle t_{2} \approx 4.79129 \; {\rm s}[/tex], the object would be on its way back to the ground and would also be [tex]5\; {\rm m}\![/tex] away from the ground.
