Question

A cat jumps from a window 3 m from ground level. Its initial speed is 3 m/s, at 30° above the horizontal. Disregard air resistance. a) How long does it stay in the air? b) How far from the building does it hit the ground? c) What is its velocity when it hits the ground?

Answer 1

Answer:

a) $t=0.95s$

b) $x=2.46m$

c) $v=8.23m/s$

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

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

$0=3+3sin(30)t-\frac{1}{2}(9.8)t^2$

Solving the quadratic equation for t:

$t=\frac{-b±\sqrt{b^2-4ac}}{2a}$

$a=-\frac{1}{2}(9.8)\\b=3sin(30)\\c=3$

$t=-0.64s$ or $t=0.95s$

Since time can not be negative, the cat stays in the air for 0.95s

b) The horizontal displacement is:

$x=v_{ox}t$

$x=(3cos(30)m/s)(0.95s)=2.46m$

c) To find its velocity when it hits the ground we need to analyze both x and y motion

$v_{x}=v_{ox}+at=3cos(30)=2.6m/s$

$v_{y}=v_{oy}-gt=3sin(30)m/s-9.8m/s^2(0.95s)=-7.81m/s$

So, v is:

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(2.6m/s)^2+(-7.81m/s)^2}=8.23m/s$