plzz solve quea no. 129
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this is related to projectile motion .
let any particle is projected with v m/s at an angle @ with horizontal .
then ,
Range =v^2sin2@/g
maximum height =v^2sin^2@/2g
according to question ,
Range =2 maximum height
v^2sin2@/g =2v^2sin^2@/2g
use sin2@ [email protected]@
v^[email protected]@/g = v^2sin^2@/g
2 =tan@
hence,
sin2@ =(2tan@)/(1+tan^2@)
=(2 x 2)/(1+4)=4/5
now,
Range = v^2sin2@/g
put sin2@=4/5
Range =4v^2/5g
hence, option (1) is correct.
let any particle is projected with v m/s at an angle @ with horizontal .
then ,
Range =v^2sin2@/g
maximum height =v^2sin^2@/2g
according to question ,
Range =2 maximum height
v^2sin2@/g =2v^2sin^2@/2g
use sin2@ [email protected]@
v^[email protected]@/g = v^2sin^2@/g
2 =tan@
hence,
sin2@ =(2tan@)/(1+tan^2@)
=(2 x 2)/(1+4)=4/5
now,
Range = v^2sin2@/g
put sin2@=4/5
Range =4v^2/5g
hence, option (1) is correct.
Anonymous:
hey! .....sin2@=2sin@[email protected] u wrote
I don't understand
mathematics formula used there
yes dear here some math tool use
sin2x = 2sinx.cosx
thnx
:-)
now I think you understand
:----)
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