Question

A stone is thrown from a point with a speed 5 m/s at an angle of elevation

Answer 1

Answer:

60 divided by 5 =12

this is the right answer

Answer 2

Complete question:

A stone is thrown from a point with a speed of 5 m/s at an elevation angle of θ. From the same point and at the same instant, a person starts running at a constant speed of 2.5 m/s to catch the stone. If the person will be able to catch the ball then, what should be the angle of projection θ?

Answer:

The angle of projection is 60°

Explanation:

Given,

The speed of stone (V) = 5 m/s

The angle at which it is thrown is θ

The speed of the person (v) = 2.5 m/s

To find,

The angle at which it is thrown (θ)

Calculation,

For a person to catch the ball they should travel the same distance along horizontal axis in time T.

Distance covered by person, R = 2.5 m/s × T...(1)

Time of flight of the ball:

$T = \frac{2V\sin(\theta)}{g}$

Substituting T in equation (1)

$R = 2.5 m/s \times\frac{2V\sin(\theta)}{g}$

$\implies R = 2.5 \times \frac{2\times 5\times \sin(\theta)}{g}$

$\implies R = \frac{25 \sin(\theta)}{g}$.......(2)

Now, the range of the stone is:

$R = \frac{V^2\sin(2\theta)}{g}\\ \implies R = \frac{25\sin(2\theta)}{g}$....(3)

equating equations (2) and (3)

We get,

sin(2θ) = sin(θ)

⇒ 2 sin(θ) cos(θ) = sin(θ)

⇒ cos(θ) = 1/2

⇒ θ = 60°

therefore,  the angle of projection (θ) is 60°.

#SPJ3