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OpenStudy (anonymous):

two seconds after the projection upwards from level ground , a projectile is moving at 30 degree to the horizontal after one more second its velocity is horizontal.the velocity of the projection was in ? (g=10m/s)

OpenStudy (anonymous):

20root3 is the answer plzz solve for it

OpenStudy (aravindg):

@shivam_bhalla , @cshalvey

OpenStudy (anonymous):

@.Sam.

OpenStudy (anonymous):

Yes @AravindG .

OpenStudy (aravindg):

pls dont help him guys he is making us do his hwork

OpenStudy (anonymous):

no u tell me the steps i will do it

OpenStudy (anonymous):

@shameer, would you like to show your attempt to solve the problem?

OpenStudy (aravindg):

dont act like u r wrking..COZ U R NOT!!

OpenStudy (anonymous):

i dont need the final answer because i know that it is 20root3

OpenStudy (anonymous):

yes i will show my work @shivam_bhalla

OpenStudy (anonymous):

Then Please show. We are waiting.

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

i think 1st i have to find this T=2usin theta/g then find u is that correct @shivam_bhalla

OpenStudy (anonymous):

@shameer1 , when does the projectile have only horizontal velocity ??

OpenStudy (anonymous):

when it is thrown horizontally/..

OpenStudy (anonymous):

\[\text{For the vertical velocity at }t_0\text{, we can use the equation for the time to reach maximum height.}\]\[t_h = 3 = \frac{v_0\sin(\theta)}{g}\]\[\implies v_0\sin(\theta) = 30 \text{m/s}\] Now if we can figure out horizontal velocity... :(

OpenStudy (anonymous):

And then use pythagorean theorem to find the velocity of projection

OpenStudy (anonymous):

hey i got it as 15root3 but it should be 20root3

OpenStudy (ajprincess):

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