Question

Two objects are thrown up at angles of 45 degree and 60 degree with the horizontal.if both objects attain the same vertical height then the ratio of the magnitude of velocities with which these are projected is

Answer 1

Explanation:

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Answer 2

Given:

$\theta=45^{\circ}$

$\theta'=60^{\circ}$

Height attained by both objects are same.

To find:

Ratio of the magnitude of velocities with which these are projected.

Solution:

We know that height attained by projectile when projected with angle $\theta$

$h=\frac{v^2sin^2\theta}{2g}$

Using the formula

$h=\frac{v^2sin^245}{2g}$

$h'=\frac{v'^2sin^260}{2g}$

$h=h'$

$\frac{v^2sin^245}{2g}=\frac{v'^2sin^260}{2g}$

$(\frac{v}{v'})^2=\frac{sin^260}{sin^245}=\frac{3}{2}$

$\frac{v}{v'}=\sqrt{\frac{3}{2}}$