Question

in case of a projectile fired at an angle equally inclined to the horizontal and vertical with velocity U the horizontal range is​

Answer 1

Answer:

The horizontal Range of a projectile is given by R= u^2sin2∆÷ g where ∆ is the agle made by projectile and g is acceleration due to gravity. Substitute the values and get the answers

Answer 2

ANSWER:

Option a). $\mathrm{u}^{2} / \mathrm{g}$

EXPLANATION:

The range of the projectile travelling has two equations both vertical and horizontal

The vertical form of motion of the projectile is stated as

$y=u t \sin a-\frac{1}{2} g t^{2}$

The horizontal form of motion of the projectile is

$x=u t \cos a$

Now there is no acceleration in the motion therefore, Acceleration (a) = 0

And the vertical motion becomes zero when hit the ground so

$y=u t \sin a-\frac{1}{2} g t^{2}$

$0=u t \sin a-\frac{1}{2} g t^{2}$

Now if the x = Range then time(t) = $2 u \sin a / g$

Then, placing time and x, we get the range as

$\frac{u^{2} \sin 2 \alpha}{g} ; \alpha=45$

$\frac{u^{2} \sin 90}{g}=\frac{u^{2}}{g}$

Therefore, the range of the object is $\mathrm{u}^{2} / \mathrm{g}$