Answer: 180.102m
Explanation:
This situation is a good example of the projectile motion or parabolic motion, in which the travel of the ball after it is thrown from the top of the cliff has two components: x-component and y-component. Being their main equations as follows:
x-component:
[tex]x=V_{o}cos\theta t[/tex] (1)
Where:
[tex]V_{o}=30m/s[/tex] is the ball's initial speed
[tex]\theta=40\°[/tex] is the angle above the horizon
[tex]t[/tex] is the time since the ball is thrown until it hits the ground
y-component:
[tex]y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex] (2)
Where:
[tex]y_{o}=150m[/tex] is the initial height of the ball
[tex]y=0[/tex] is the final height of the ball (when it finally hits the ground)
[tex]g=9.8m/s^{2}[/tex] is the acceleration due gravity
Knowing this, let's begin with the calculations:
Firstly, we have to find the time the ball elapsed traveling. So, we will use equation (2) with the conditions given above:
[tex]0=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex] (3)
Isolating [tex]t[/tex] from (3):
[tex]t=\frac{V_{o}}{g}(sin\theta+\sqrt{{sin\theta}^{2}+2\frac{g.y_{o}}{V_{o}^{2}}})[/tex] (4)
[tex]t=7.83s[/tex] (5)
Substituting (5) in (1):
[tex]x_{max}=30m/s.cos(40\°) (7.83)[/tex] (6)
Finally:
[tex]x_{max}=180.102m[/tex] (7)
