A cat is chasing a mouse as shown in the figure. The mouse runs horizontally with a speed 10m//s.The cat runs with a constant speed of 20m//s. At what angle to the horizontal should the cat run in order to catch the mouse?

A cat is chasing a mouse as shown in the figure. The mouse runs horizo

1.	A cat is chasing a mouse as shown in the figure. The mouse runs horizontally with a speed 10m//s.The cat runs with a constant speed of 20m//s. At what angle to the horizontal should the cat run in order to catch the mouse?
Answer» Solution :Method-1[Analysis of velocity components] Assume that the cat and the mouse do meet at point .D. as shown. Let D be the horizontal point above .D. . The cat has a velocity .v. which carries it to point .D. in time .t.. Since the velocity is inclined, we can resolve it to component velocities `v_(x),v_(y).v(x)` carries cat from point C to D in time t and `v_(y)` carries cat from C to D in t. We thus SEE that theX-component of velocity, `v_(x)` is responsible for keeping cat directly above mouse . Y- component of velocity vrings cat nearer and nearer tp mouse. HENCE the cat should run such that `v_(x)=10`. `:.20 cos alpha =10 ` ` cos alpha=1//2,alpha =60^(@)` only when cat runs at this angle it will remain at the same level as the mouse at all times. to find teh time at which they meet , we analyze motion along Y-axis . the cat has a Y-component of velocity of `20 sin 60=10sqrt3=17.32m//s`. It has to TRAVEL a distance of 173m in `-ve` Y direction . Hence time taken `t=("VERTICAL distance" )/("vertical component velocity" )=(173)/(17.3)=10sec`.