Hey! Goodmorning brainliacs..Answer this Physics question...☺️☺️☺️
A projectile is projected at an angle @ (>45°) with an initial velocity u. The time at which its horizontal component will be equal to the vertical component in magnitude? (@ stands for alpha)
Options are in the picture.
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Answers
i cant snd a pic now so i will try to xplain here.
Firstly make a rough diagram of the motion.
ux=ucos@ uy=usin@
u know that x component of velocity always remains constant so work with the y component.In y direction applying NLM,(uy will bcm equal to ucos@ as per qstion)
vy=uy-gt >> ucos@=usin@ -gt
>>gt=usin@-ucos@
>>t=u/g * (sin@-cos@)
option C i guess.
option B is also correct.
we know that in a projectile motion the particle passs through a point at a height twice during the whole motion. A max. height we know thatby component of velocity bcms 0. so when it comes down again the y component of velocity will again bcm equal to ucos@ at sm point but this time u need to use the minus(-) sign in the soln above as the y component pf velovcity is pointed downwards.hope it helped u!!