Question

A ball is thrown up at an angle 30° with the horizontal. If R and H represent its horizontal range and
maximum height reached, then
1) R = 2H
2) R = H
3) R = H/2
4) R = 4 square root 3 H​

Answer 1

Answer:

let a be the angle of projection

R= u²sin 2a/g u = velocity g = gravity

H= u²sin²a/2g

Now it is given the angle is 30 so substitute the following

R = u²sin(2*30)/g = u²sin 60/g =√3/2u²/g ---(1)

sin 60= √3/2 sin 30= 1/2

H= u² (sin 30)²/2g. = u² (1/2)²/2g. = 1/4u²/2g = H--(2)

now rearrange H;

multiply and divide eqn (2) by √3 ,we get;

H= √3/4√3 * 1/2 u²/g

H= 1/4√3 * √3/2u²/g ----(3)

Substitute (1) in (3)

H= 1/4√3 R

by rearranging;

R= 4 square root 3 H

which is the answer

it's option 4

Answer 2

The correct option is 4) R = 4 square root 3 H.

Given:

A ball is thrown up at an angle of 30° with the horizontal. R and H represent its horizontal range and maximum height reached.

To Find:

The relation between R and H.

Solution:

To find the relation between R and H we will follow the following steps:

As we know,

$R = {u}^{2} \frac{sin2x}{g}$

$H = \frac{ {u}^{2} \times sinx }{2g}$

Here, x is the angle = 30°

Now,

$R = {u}^{2} \frac{sin60}{10} = \frac{ \sqrt{3} }{20} {u}^{2}$

$H = \frac{ {u}^{2} \times {(sin30)}^{2} }{20} = \frac{1}{80} {u}^{2}$

$As sin60 \: = \frac{ \sqrt{3} }{2} and \: sin30 = \frac{1}{2}$

Taking ration of R and H we get,

$\frac{R }{ H } = \frac{ \sqrt{3} \times 80 \times {u}^{2} }{20 \times 1 \times {u}^{2} } = 4 \sqrt{3}$

$R = 4 \sqrt{3} H$

Henceforth, the correct option is 4) R = 4 square root 3 H.

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