Physics, asked by AMANSIDDIQUI4068, 1 year ago

A projectile is projected with initial velocity (6i+8j) m/sec. If g=10 find horizontal range

Answers

Answered by tripathishashank872
20

The velocity of the projectile is given in the vector form . That is , 6i+8j. We will find out the magnitude of this vector by the given way
√(6¡)^2+ (8j)^2 = √100=10
So the velocity of projectile comes out to be 10m/s.
Now we have the x component of the projectile equal to 6 . Therefore 10cosx=6 . cosx=3/5
By this we find sinx=4/5(by Pythagoras theorem )
So range is equal to (u^2sin2x)/g
Since sinx=4/5 sin2x will be equal to 2sinxcosx( trigonometric identity . ) . sin2x=2(4/5)(3/5)= 24 / 25
Now put this in the formula
u^2sin2x/g=100(24/25)/10= 9.6m
I hope it is the correct answer.

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