Posted by Tobin on Thursday, August 27, 2009 at 2:00am.
Tom's horizontal velocity component will remain 5.0 m/s. Only the vertical component will change. The time T required to fall a vertical distance Y can be computed from the formula
Y = (g/2)T^2 = 1.5 m
which leads to
T = sqrt(2Y/g) = sqrt(3.0/9.8)
Calculate that number.
The horizontal distance from the edge of the table where Tom lands is
X = 5.0 m/s * T
a). With the above assumption, initial velocity, just off the table surface is u = 4.8 m/s (assume in the horizontal, X direction)
Vertical height, distance, S = 1.0 m (given)
Acceleration = acceleration of gravity = 9.8 m/s2 (assumed).
To calculate the horizontal distance traveled, first calculate the time of fall, t.
Formula to be used, S = ut + 1/2 a t2
For vertical component of the motion, S= y = 1.0 m; u = Uy = 0.
So, y = 0 +4.9 t2; t = sq-rt y/4.9 = sq-rt 1/4.9 = 0.452 seconds.
Now consider the horizontal component of motion.
Ux = u = 4.8 m/s, a = ax =0; S = Sx =?
Applying the same formula, S = X = ut = 4.8x0.452 = 2.1696 m.
This may be approximated to 3 significant figures.
Then, X = 2.17 m.
The cat will land on the floor, 2.17 m from the edge of the table(measured along the floor).
2(b). Velocity component. Since we assume, the acceleration is only due to gravity, it is along the vertical direction and its horizontantal component is zero. So, ay = -9.8 m/s2 (negative for downward, as we assume up is positive).
So, the horizontal component of the motion is uniform; the velocity remains constant. vx = ux = u =4.8 m/s.
The vertical component: Initial velocity, uy = o;
Using the formula v = u +at;
vy = 0 + -9.8x0.452 = - 4.4296 m/s.
Again, approximating to 3 sf, vy = -4.43 m/s.
The negative sign shows, it is downward.
(we assumed, up is +ve).
thanks