Question

ratio of their times of flight and ratio of ranges and ratio of maximum heights

Answer 1

HELLO THERE!

Let's analyse the given data:

The initial velocities is same in each case, say, it is u.

In the first case, the ball is projected at 30° from horizontal and in the second case, it is projected at 60° from the horizontal.

Time of flight of the first ball:

$T_{1} = \frac{2usin\theta}{g}\\\\=\frac{2usin30}{g}\\\\= \frac{u}{g}$

(where u is the initial velocity with which it is thrown, and g is the acceleration due to gravity).

Time of flight of the second ball:

$T_{2} = \frac{2usin\theta}{g}\\\\= \frac{2usin60}{g}\\\\= \frac{\sqrt{3}u}{g}$

Now, ratio is:

$\frac{T_{1}}{T_{2}} = \frac{\frac{u}{g}}{\frac{\sqrt{3}u}{g}}\\\\= \frac{1}{\sqrt{3}}$

Hence, the ration of the time of flights is:

1 : √3

Let's check the ratio of their ranges too!

Range of the first ball:

$R_{1} = \frac{u^{2}sin2\theta}{g} \\= \frac{u^{2}sin60}{g} \\= \frac{\sqrt{3}u^{2}}{2g}$

Range of the second ball:

$R_{2} = \frac{u^{2}sin2\theta}{g} \\= \frac{u^{2}sin120}{g} \\= \frac{\sqrt{3}u}{2g}$

So, ratio of the ranges:

$\frac{R_{1}}{R_{2}} = \frac{\frac{\sqrt{3}u}{2g}}{\frac{\sqrt{3}u}{2g}} \\= 1:1$

So, ratio of their ranges is 1:1.

You can even calculate this without calculations. We know, that ranges of two particles is same, if they are thrown with the same velocity, and angles complementary to each other.

The first ball is thrown with 30° and the second with 60°. These are complementary angles, so the ratio of their ranges is 1:1.

THANKS!