Physics Homework

Question

A running mountain lion can make a leap 10.0 m long, reaching a maximum height of 3.0 m.

a) What is the speed of the mountain lion just as it leaves the ground?

b) At what angle does it leave the ground?

Answer 1

Let h=h₀ +v_0yt -at²/2 where h₀ =0 and h=3m v= v_0y -at and at peak v=0 so v_0y =at and t= v_0y/a. Here a=9.8 m/s² and v_0y = vsinθ so t=9.8/vsinθ using this 3 = 0 + v²sin²θ/9.8 -4.9v²sin²θ/9.8² simplifying this equation gives 3=v²sin²θ/9.8 - 4.9•2v²sin²θ/2•9.8² and 3= v²sin²θ/9.8 - v²sin²θ/ 19.6 = v²sin²θ/19.6 which can be written as 6=v²sin²θ/9.8 also v_0x= vcosθ and 10= vcosθ•2vsinθ/9.8 and 49= v²cosθsinθ but v² =9.8•6/sin²θ so 49=58.8ctnθ and tanθ =58.8/49 using the inverse tan θ= 50.19º as the angle at which the mountain leaps and v² =58.8/.59 which gives velocity of v = 9.98m/s as the initial velocity of departure.