Question

Tom the cat is chasing Jerry the mouse across
a table surface 1.7 m off the floor. Jerry steps
out of the way at the last second, and Tom
slides off the edge of the table at a speed of
4.2 m/s.
Where will Tom strike the floor? The ac-
celeration of gravity is 9.8 m/s2 .
What speed will Tom have just before he hits?
Answer in units of m/s

Answer 1

Let's break this up into vertical and horizontal components.

Vertical

At the top of the table, the vertical component of velocity is zero because Tom has not yet begun to descend. So, \(v_{ov}=0\). gravity begins to accelerate Tom and he hits the floor with vertical velocity \(v^2_{fv}=2\times g\times h\), where h is 1.7 m, the distance of the table to the floor. From this, we get that final vertical velocity the Tom hits the floor with is \(v_{fv}=5.8~m/s\).

Horizontal

To determine how long he was falling, we use \(v_{fv}\times g \times t\), where t is the time in flight. From this, we get that \(t=0.59~s\).

Since, Tom has been traveling for 0.59 seconds, his horizontal distance traveled is \(v_{v}=g\times t\), where \(v_{v}\) is Tom's horizontal velocity, which stays constant because he is not accelerating or decelerating in this direction. From this, we get that \(v_{v}= 2.5~m\).

Total Velocity

His total velocity will include his horizontal and vertical speeds: \(v_{total}=\sqrt{5.8^2+4.2^2}=7.2~m/s\).

Let me know if you have any questions.

Answer 2

Then, you should multiply t to Vx.

Then you should find also Vy by using Vy(fin)=Vy(in) +g*t

Tom's speed when it hits the ground V=(Vx^2 +Vy^2)^(1/2)