Question

A body projected at 45° with a velocity of 20 m/s has a range of 10mThe decrease in range due to air resistance is
(g = 10 ms-1)
(A) 0
(B) 10 m
(C) 20 m
(D) 30 m​

Answer 1

During projectile motion the horizontal distance  covered by the body is called horizontal range .

Say horizontal range belongs to x axis and horizontal distance is S .

Then R(S) = u cos theta × T (time of flight)

We know that , time of flight = 2 u sin theta ÷ g

So , R = u cos theta × 2 u sin theta ÷g = u^2 sin 2 theta ÷g . ………………..(1)

Now according to the question , theta =45 °and u= 20 m/s . g = 10 m/s .

Putting this values in number 1 equation ,

R= 20×20× sin 90°÷10 =40 m [ as sin 90° =1]

= 40 m

Now subtract this range from the given range

40 - 10 = 30 m

answer is 30m

Answer 2

$range = \frac{u {}^{2}sin2 \alpha }{g}$

With the help of this formula

Range = 40m without Air risistance

reduction in range due to air = (range without air resis)-(range with air resis)

Reduction =40-10

=30m

I think it will helo you