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A projectile is thrown at an angle and another is thrown at (90- ) from the same point both with the velocities 78.4ms-1. The second reaches 36.4m higher than the first. Find the individual heights.
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Solution:

Given,

Angle of projection of first projectile, =θ

Angle of projection of second projectile, =(90θ)

Velocity of projection, u=78.4 ms1

and, H2=H1+36.4 m

H2H1=36.4 m

Where, H1 is the maximum height of first one and H2 is the maximum height of second one.

H1=u2sin2θ2 g and , H2=u2sin2(90θ)2 g=u2cos2θ2 g

Adding, H2+H1=u2cos2θ2 g+u2sin2θ2 g

=u22 g=78.4×78.42×9.8=313.62H2=36.4+313.6=350H2=175 m

Also, H1=H236.4=17536.4

H=138.6 m

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