Hi there! Have questions about your homework? At Brainly, there are 60 million students who want to help each other learn. Questions are usually answered in less than 10 minutes. Try it for yourself by posting a question! :D

Hi there! Have questions about your homework? At Brainly, there are 60 million students who want to help each other learn. Questions are usually answered in less than 10 minutes. Try it for yourself by posting a question! :D

First of all find the time in horizontal direction for motion in horizonrtal direction initial velocit,u=ucosx acceleration,a=0 distance,S=x time,t=t so,by second law of motion x=ut+1/2*a(t*t) x=ucosx.t+0 t=x/ucosx then, for motion in vertical direction initial velocity,u=usinx acceleration,a=g distance,S=y time,t=t similarly by second law of motion y=ut+1/2*a*(t*t) y=usinx.t+1/2*g(t*t) putting value of t from above y=usinx.(x/ucosx)+1/2*g*(x/ucosx*x/ucosx) finally you'll get y=utanx*x+g/2(u*u)*cos*cosx*(x*x) and this is a equation of parabola

Mow the time of flight in air means motion in vertical direction therefore,In vertical direction time,t=Total time /2 acceleration,a=-g final velocity,v=0 initial velocity,u=usinx then, by first law of motion v=u+at 0=usinx-g*T/2 Ultimatily, Time of flight is,T=2usinx/g