Tom and Jerry

After t seconds the mouse has moved 0.5t metres away from the cat. At the instant the cat leaps with a vertical velocity $2/ \sqrt{2}$ m/s and so the total time of flight is $2\times 2/ \sqrt{2} \times 9.8= 0.288$ seconds. The horizontal velocity of the cat with respect to the mouse is $2/ \sqrt{2}-0.5=0.914$ m/s. The total distance covered with respect to the mouse is 0.5t which is equal to velocity times time of flight= $0.914 \times 0.288=0.263$ metres which gives $t= 0.263 \times 2= 0.528$ seconds.

horizontal distance traveled by mouse=horizontal distance traveled by cat let the cat wait for a time=t horizontal speed of mouse=.5 angle of leap=45 degree with horizontal initial speed of cat(u)=2m/s horizontal range covered by cat=\frac {u cos45(2 (\sqrt{2})}{9.8} time taken to cover this horizontal range=\frac {2 (\sqrt{2})}{9.8} as horizontal distance traveled by mouse=horizontal distance traveled by cat .5 (t+\frac {2 (\sqrt{2})}{9.8})=\frac {u cos45(2 (\sqrt{2})}{9.8}
by rearranging the terms and substituting value of u=2 we get t=(\frac {2
cos45(2 (\sqrt{2})}{9.8}-.5 (\frac {2 (\sqrt{2})}{9.8})) 2 this is=0.52771151788304182677516556648782

The velocity of the mouse=0.5 m/s(u) The cat waits for time t seconds...... now the cat leaps with an angle of 45 degrees and directly lands on the mouse.. So lets put the logic of physics in this story, as the cat jumped on the mouse it gave itself a velocity of 2 m/s and then was in the influence of gravity,which mean he was in projectile motion. As the cat directly landed on the mouse its sure that:- range covered by cat=distance moved by the mouse....equation obtained. now distance move by rat took a time=t+time of flight of cat =t+2usin45/g(time of flight =2usinx/g) substitute the values =t+0.28 now distance moved by rat=velocity*time =0.5(t+0.28) now the last nail in coffin,just equate it with range of cat. 4/9.8=0.5(t+0.28).............(range=u^2sin2x/g)and(with 45 degree we get max range of u^2/g) just solve this equation to get t=0.52

The total time the cat is in the air, $t_1$ is given by $y_f=y_i+v_i (sin \theta) t_1-\frac{1}{2}gt_1^2\rightarrow t_1=\frac {2 v_i sin \theta}{g}$ . The x position of the cat upon landing is then $x_c=v_i cos \theta t_1$ . The x-position of the mouse is $x_m=v_m(t_1+t)$ . Since the cat catches the mouse, $x_c=x_m$ and so we can solve for t.

Consider vertical component of cat's motion.The cat reach its maximum point (v=0) after a time p. v=u+ap 0= 0.5 sin 45-9.8p p=0.03608 s

Since the motion is symmetrical, hence the travelling time of cat would be 2p=0.07216 s.

Consider the horizontal component of cat's motion. The travelling distance of cat can thus be determined from s=u x 2p=2 cos45 x 0.07216 =0.1020 m

The time taken for mouse's motion is thus T=s/U=0.1020/0.5= 0.2040 s

Since the mouse and the cat took time T and 2p respectively to reach the same point, therefore t=T-2p=0.2040-0.07216 =0.13184 s

First we must consider the cat's motion in 2-dimension, and use components to simplify it.

The first thing we must think about is the cat's starting point and ending point. The question states that the cat jumps from the ground and lands back on the ground , so from that we can say that the cat's overall change in height is 0, seeing as how he starts at h=0 and ends at that same height.

Now, we can use this information to find a total time for the cat's jump using the equation:

$d = v_i \cdot t + \frac {1}{2} a_{av} \cdot t^2$

Using components, we change it to the y-components of the equation and solve for time... where $d_y = 0 , v_{iy} = 2 \times Sin45$ , and $a_y = -9.8$

Using quadratic formula to solve the quadratic we can get the time 0.289 seconds. This represents the time the cat takes for his jump to be completed. Using this time, we can find the range, or displacement in the x using the same equation, in the x-component...

In the x-component form of the equation, we know there is no acceleration in the x-component, in projectile motion. So, the final term is 0, then we just solve for d_x ....

$d_x = v_{ix} \cdot t$ - where we get the range 0.408m

This represents the distance in which the cat travels in the x-direction in the time of this jump.

Now for the mouse's motion, a more simple problem. We know that the cat lands on top of the mouse, which means their horizontal (x) displacement must be the same at the time of the cat's landing, so using the equation for uniform velocity, we plug in the cat's range into the equation to solve for a time...

$v_x = 0.5 m/s = d / t$ --> d= 0.408m

$t = d / v_x$ = 0.816 seconds.

Now we have the time which it takes the mouse to move that distance. Now all we have to do is simply find the change in time from the cat's leap and the mouse's run across the floor and we can see the gap in time the cat waited to jump...

$t_m - t_c = t_w$ .... where $t_w$ is the time waited by the cat... from that we get:

$t_m - t_c = t_w$ = 0.527 seconds waited by the cat.