Question

a projectile is thrown at some angle with the horizontal. when it is at the highest point of its motion , the speed is 1 by root 2 times the speed of the projection. the angle of the projection with the horizontal is​

Answer 1

Answer is given in the pic.

Answer 2

Given:

A projectile is thrown at some angle with the horizontal. when it is at the highest point of its motion , the speed is (1/√2) times the speed of the projection.

To find:

Angle of projection with the horizontal.

Calculation:

Let us assume that the initial velocity of projection is "u" and angle of Projection be $\theta$.

Then , velocity of that object at the highest point of the projectile trajectory will be ;

$\boxed{ \rm{v_{(highest \: pt)} = u \cos( \theta) }}$

Now , the question says that velocity at the highest point is 1/√2 times the initial velocity.

$\therefore \rm{v_{(highest \: pt)} = \dfrac{1}{ \sqrt{2} } (u)}$

$= > \rm{u \cos( \theta) = \dfrac{1}{ \sqrt{2} } (u)}$

$\rm{ = > \cos( \theta) = \dfrac{1}{ \sqrt{2} } }$

$\rm{= > \theta = {45}^{ \circ} }$

So, final answer is:

$\boxed{ \bf{ \theta = {45}^{ \circ} }}$