Question

Two bodies are projected with same speed of 40ms at angles (45-) and (45+). The ratio
their range is​

Answer 1

Answer:

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Answer 2

Given:

Two bodies are projected with the same speed of 40ms at angles (45-) and (45+).

To Find:

The ratio their range.

Solution:

To find the ratio of their range. we will follow the following steps:

As we know,

The formula for finding the range is

$\frac{ {uo}^{2}sin2θ }{g}$

Here,

R= horizontal range in metre.

u0 = initial velocity in metres per second.

g = acceleration due to gravity= 9.8 ms-2.

θ = angle of the initial velocity from the horizontal plane.

Now, taking the range ratio of both the bodies we get,

$\frac{ {uo}^{2}g \: (sin2θ)1 }{{uo}^{2}g(sin2θ)2}$

uo is equal for both the particle.

So, by cancelling uo and g we get,

$= range \: ratio = \: \: \frac{(sin2θ)1}{(sin2θ)2}$

Now, put values of θ of the first and second particle.

$Range \: ratio = \frac{sin2 \times -45}{sin2 \times ( 45)} = \frac{sin-90°}{sin90°} =\frac{-sin90°}{sin90°}$ = 1:1

Since range cannot be negative so, it is its ratio is taken positively as we know a negative sign indicates the projectile is fired in opposite direction with the same angle.

Sin90° is in the first quadrant so its value is +1.

Henceforth, the ratio of their range is 1:1