Question

191. A bullet is fired from a cannon with velocity
500 m/s. If the angle of projection is 15° and
g - 10m/s? Then the range is :-
(1) 25 x 10%
(2) 12.5 x 103 m
(3) 50 x 102
(4) 25 x 102 m​

Answer 1

Answer

Option (2)

Range = 12.5 × 10³ m

Given

A bullet is fired from a cannon with velocity  500 m/s. If the angle of projection is 15° and  g = 10 m/s²

To Find

Range

Formulas to be noted

$\rm \bullet \;\; T=\dfrac{2usin\theta}{g}\\\\\rm \bullet \;\; H=\dfrac{u^2sin^2\theta}{2g}\\\\\rm \bullet \;\; R=\dfrac{u^2sin2\theta}{g}\\\\\rm where,R\ denotes\ Horizontal\ range$

Solution

Initial velocity , u = 500 m/s

Angle of projection , θ = 15°

Gravity , g = 10 m/s²

Apply formula for range ,

$\to \rm R=\dfrac{u^2sin2\theta}{g}\\\\\to \rm R=\dfrac{(500)^2sin(2\times 15^0)}{10}\\\\\to \rm R=\dfrac{500\times 500\times sin30^0}{10}\\\\\to \rm R=500\times 50\times \dfrac{1}{2}\\\\\to \rm R=500\times 25\\\\\to \rm R=12500\ m\\\\\to \rm R=12.5\times 10^3\ m$

So , Option (2) is correct

Answer 2

Explanation:

option 2 is correct....