Answer:
The projectile strikes the tower at a height of 354.824 meters.
Explanation:
The projectile experiments a parabolic motion, which consist of a horizontal motion at constant speed and a vertical uniformly accelerated motion due to gravity. The equations of motion are, respectively:
Horizontal motion
[tex]x = x_{o}+v_{o}\cdot t \cdot \cos \theta[/tex]
Vertical motion
[tex]y = y_{o} + v_{o}\cdot t \cdot \sin \theta +\frac{1}{2} \cdot g \cdot t^{2}[/tex]
Where:
[tex]x_{o}[/tex], [tex]x[/tex] - Initial and current horizontal position, measured in meters.
[tex]y_{o}[/tex], [tex]y[/tex] - Initial and current vertical position, measured in meters.
[tex]v_{o}[/tex] - Initial speed, measured in meters per second.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]t[/tex] - Time, measured in seconds.
The time spent for the projectile to strike the tower is obtained from first equation:
[tex]t = \frac{x-x_{o}}{v_{o}\cdot \cos \theta}[/tex]
If [tex]x = 600\,m[/tex], [tex]x_{o} = 0\,m[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]\theta = \frac{\pi}{4}[/tex], then:
[tex]t = \frac{600\,m-0\,m}{\left(120\,\frac{m}{s} \right)\cdot \cos \frac{\pi}{4} }[/tex]
[tex]t \approx 7.071\,s[/tex]
Now, the height at which the projectile strikes the tower is: ([tex]y_{o} = 0\,m[/tex], [tex]t \approx 7.071\,s[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex])
[tex]y = 0\,m + \left(120\,\frac{m}{s} \right)\cdot (7.071\,s)\cdot \sin \frac{\pi}{4}+\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (7.071\,s)^{2}[/tex]
[tex]y \approx 354.824\,m[/tex]
The projectile strikes the tower at a height of 354.824 meters.
