Answer:
The toy rocket will do an horizontal distance of 27.793 meters before reaching the ground.
Explanation:
This rocket experiments a two-dimension motion consisting in a combination of vertical free-fall and a horizontal uniformly acclerated motion. Flight time is governed by vertical movement and can be found by using this formula:
[tex]y = y_{o}+v_{o,y}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}[/tex]
Where:
[tex]y_{o}[/tex] - Initial height, measured in meters.
[tex]v_{o,y}[/tex] - Initial vertical speed, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]y[/tex] - Current height, measured in meters.
If we know that [tex]y_{o} = 30\,m[/tex], [tex]v_{o,y} = 0\,\frac{m}{s}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex] and [tex]y = 0\,m[/tex], then resulting polynomial is solved:
[tex]0\,m = 30\,m +\left(0\,\frac{m}{s} \right)\cdot t +\frac{1}{2} \cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}[/tex]
[tex]30 - 4.905\cdot t^{2} = 0[/tex]
The time taken by the toy rocket is:
[tex]t \approx 2.407\,s[/tex]
Now, the horizontal distance can be found by integrating acceleration function twice. That is:
[tex]v(t) = \int {a(t)} \, dt[/tex]
[tex]v(t) = 1.60\int {t} \, dt[/tex]
[tex]v(t) = 0.80\cdot t^{2}+v_{o}[/tex]
[tex]s(t) = \int {v(t)} \, dt[/tex]
[tex]s(t) = \int {(0.80\cdot t^{2}+v_{o})} \, dt[/tex]
[tex]s(t) = 0.80\int {t^{2}} \, dt + v_{o}\int\, dt[/tex]
[tex]s(t) = 0.267\cdot t^{3} + v_{o}\cdot t + s_{o}[/tex]
If we know that [tex]v_{o} = 10\,\frac{m}{s}[/tex] and [tex]s_{o} = 0\,m[/tex], then the horizontal position formula is:
[tex]s(t) = 0.267\cdot t^{3}+10\cdot t[/tex]
Where:
[tex]s(t)[/tex] - Horizontal position, measured in meters.
[tex]t[/tex] - Time, measured in seconds.
Now, the horizontal distance before reaching the ground is found: ([tex]t \approx 2.407\,s[/tex])
[tex]s(2.407\,s) = 0.267\cdot (2.407\,s)^{3}+10\cdot (2.407\,s)[/tex]
[tex]s (2.407\,s) = 27.793\,m[/tex]
The toy rocket will do an horizontal distance of 27.793 meters before reaching the ground.
