Question

Q.12
A body is projected with a velocity of 25m/s making on an angle of 53° from the horizontal.
Then, velocity of the body at the highest point will be
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Answer 1

Answer:

Given,

Projected velocity, v

angel, θ

Velocity at any time t

V

t

=vcosθ

i

^

+(vsinθ−gt)

j

^

tanα=

vcosθ

vsinθ−gt

(at t=1)tanα=

vcosθ

vsinθ−g

Angle at time 1 sec from start α=tan

−1

(

vcosθ

vsinθ−g

)

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