Answer:
a) [tex]t=0.95s[/tex]
b) [tex]x=2.46m[/tex]
c) [tex]v=8.23m/s[/tex]
Explanation:
From the exercise we know that the initial height is 3m and the initial velocity is 3 m/s at 30º above the horizontal
a) To calculate how long does the cat stays in the air we need to know that at that moment y=0
[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^2[/tex]
[tex]0=3+3sin(30)t-\frac{1}{2}(9.8)t^2[/tex]
Solving the quadratic equation for t:
[tex]t=\frac{-b±\sqrt{b^2-4ac}}{2a}[/tex]
[tex]a=-\frac{1}{2}(9.8)\\b=3sin(30)\\c=3[/tex]
[tex]t=-0.64s[/tex] or [tex]t=0.95s[/tex]
Since time can not be negative, the cat stays in the air for 0.95s
b) The horizontal displacement is:
[tex]x=v_{ox}t[/tex]
[tex]x=(3cos(30)m/s)(0.95s)=2.46m[/tex]
c) To find its velocity when it hits the ground we need to analyze both x and y motion
[tex]v_{x}=v_{ox}+at=3cos(30)=2.6m/s[/tex]
[tex]v_{y}=v_{oy}-gt=3sin(30)m/s-9.8m/s^2(0.95s)=-7.81m/s[/tex]
So, v is:
[tex]v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(2.6m/s)^2+(-7.81m/s)^2}=8.23m/s[/tex]
