Question

A cat jumps off a piano that is 1.3m high. The initial velocity of the cat is 3m/s at an angle of 37degrees above the horizontal. How far from the edge of the piano does the cat strike the floor.

Answer 1

Answer:

$x=1.75m$

Explanation:

From the exercise we have that

$y_{o}=1.3m\\v_{o}=3m/s, \beta =37$

To find how far from the edge of the piano does the cat strike the floor, we need to calculate its time first

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

At the end of the motion y=0m

$0=1.3+3sin(37)t-\frac{1}{2}(9.8)t^{2}$

Solving for t

$t=-0.36 s$ or $t=0.73s$

Since the time can't be negative the answer is t=0.73

Knowing that we can calculate how far does the cat strike the floor

$x=v_{ox}t=3cos(37)(0.73)=1.75m$