Question

A body is thrown with velocity (4i+3j) metre per second.Its maximum height

Answer 1

Explanation:

Assuming SI units, the x-component of velocity is 3m/s and y-component is 4m/s .

The angle of projection is θ=tan−1(4/3)

The formula for max height H is:

H=u2sin2θ / 2g

The formula for max range R is:

R=u2sin2θ/g

From the two equations we get the relation between H , R and θ

H=Rtanθ/4

If we replace R with 2R(Range is doubled) while keeping θ constant, we see that the maximum height gets doubled. The angle is not modified as the question does not say anything about changing angles.

Answer 2

Explanation:

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