Question

A body projected at an angle of 45 with velocity 20m/s has range of 10m. The decrease in range due to air resistance

Answer 1

Answer:

The answer is 30 m.

Explanation:

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 2

30

R=$\frac{2u \sin(45) }{g} \times u \cos(45)$

$\frac{20}{ \sqrt{2} } \times \frac{4}{ \sqrt{2} }$

$\frac{80}{2 \: } = \: 40$

$so \: total \: range \: is \: 40 \:$

$sub \: the \: given \: range \: from \: 40$

$40 - 10 = 30$