Answer:
a) [tex]v_{o}=16.6m/s[/tex]
b) [tex]v=14.6m/s[/tex]
[tex]\beta =-55.32º[/tex]
Explanation:
From the exercise we know that the ball is thrown at an given angle, strikes the building 18.0 m away, that will be our horizontal displacement. Also, we know whats the final height of the ball which is 8.0 m
a) Given the information that we have we can calculate the initial velocity by solving the following formula
[tex]x=v_{o}cos(60)t[/tex]
Since we don't know how much time does the ball take to strike the building we need to calculate that first
[tex]v_{o}=\frac{x}{tcos(60)}=\frac{18m}{tcos(60)}[/tex] (1)
If we analyze the vertical displacement
[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^2[/tex]
[tex]8m=\frac{18m}{tcos(60)}sin(60)t-\frac{1}{2}(9.8m/s)t^2[/tex]
[tex](4.9m/s^2)t^2=(18m)tan(60)-8m[/tex]
[tex]t=\sqrt{\frac{(18m)tan(60)-8m}{4.9m/s^2} }=2.17s[/tex]
Knowing that time we can calculate the ball's initial velocity from (1)
[tex]v_{o}=\frac{18m}{(2.17s)cos(60)} =16.6m/s[/tex]
b) To calculate the magnitude and direction of the ball's velocity we need to find the x and y components of velocity
[tex]v_{x}=v_{ox}+at=(16.6m/s)cos(60)=8.3m/s[/tex]
[tex]v_{y}=v_{oy}+gt=(16.6m/s)sin(60)-(9.8m/s^2)(2.17s)=-12m/s[/tex]
The magnitude of velocity is:
[tex]v=\sqrt{v_{y}^{2}+v_{x}^2}=\sqrt{(-12m/s)^2+(8.3m/s)^2}=14.6m/s[/tex]
The direction of the ball's velocity is:
[tex]\beta =tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-12m/s}{8.3m/s})=-55.32º[/tex]
