Projectile brain teaser

yes :)

How did u get 1.2 s ?? time period is 1 s

oops lol i thought 12m/s :P

total time period = 1s. At t=1s, then the angle will be 30^..

Also since angle is 30^., there will never be such a case as the max angle is 30^. while landing and it will not reach 60^. which is needed to attain the given condition.

this is ur favorite question i supppose..:P

Now let me explain the entire thing.

U = ucos(θ) i + usin(θ) j

V= ucos(θ) i + (usin(θ) - gt) j

if U _|_ V then U . V = 0

Whence ,

u² - gtusin(θ) = 0
≡ t = u/(gsin(θ))

But this t < T (time of flight)

u/(gsin(θ)) ≤ 2usin(θ)/g

θ ≥ 45° limiting condition.

Infact , at θ = 45° the landing angle is also 45° and the perpendicularity of velocities are achieved at the point of landing. If θ < 45° this condition will never be achieved.

u=5√3i+5j
v=(5√3i=5j)-10jt

u.v=0
=>(5√3i+5j)[(5√3+5j)-10jt]=0

solving,
t=2

how can u get this aditya solved it right
{u2 - gtusin(θ) = 0
≡ t = u/(gsin(θ))
}
This step i didnt understand can any one explain it to me??

we can also change the co-ordinate axis nd then easily get the answer....

Its easy...the velocity will b perpendicular only wen it wil complete its projectile....i.e., the time perios = 2usin30/10
which is equal to 1 s .....

best idea is to use u . v = 0

and I like a dumbo went on with the whole calculation.. without focusing on the angle..