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

`x=1.75m`

Step-by-step 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+(1)/(2)gt^(2)`

At the end of the motion y=0m

`0=1.3+3sin(37)t-(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`