Answer:
a) vertical position at 2s: 50.68 m
b) horizontal position at 2s: 13.05 m
Explanation:
This situation is a good example of the projectile motion or parabolic motion, in which the travel of the rock has two components: x-component and y-component. Being the equations to find the position as follows:
x-component:
[tex]x=V_{o}cos\theta t[/tex] (1)
Where:
[tex]V_{o}=7.2 m/s[/tex] is the rock's initial speed
[tex]\theta=25\°[/tex] is the angle at which the rock was thrown
[tex]t=2 s[/tex] is the time
y-component:
[tex]y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex] (2)
Where:
[tex]y_{o}=25 m[/tex] is the initial height of the rock
[tex]y[/tex] is the height of the rock at 2 s
[tex]g=9.8m/s^{2}[/tex] is the acceleration due gravity
Knowing this, let's begin with the answers:
a) Vertical position at 2 s:
[tex]y=25 m+(7.2 m/s)sin(25\°) (2 s)-\frac{(9.8m/s^{2})(2)^{2}}{2}[/tex] (3)
[tex]y=25 m+6.085 m-19. 6 m[/tex] (4)
[tex]y=50.68 m[/tex] (5) This is the vertical position at 2 s
b) Horizontal position at 2 s:
[tex]x=(7.2 m/s) cos(25\°) (2 s)[/tex] (5)
[tex]x=13.05 m[/tex] (6) This is the horizontal position at 2 s
