Question

When a projectile is thrown up at an angle, to the ground, the time taken by it to rise and fall are related as?

Answer 1

Hint: Whenever an object is thrown, in air or from air, it moves in a plane only and covers a distance in that plane only. One such motion is Projectile motion. In a projectile motion, one of the most important conditions is the uniform gravitational field of the earth (in case if we are studying the projectile motion on earth).

Formula Used:
The general equation of ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\forall a > b$

Complete step by step solution:
A projectile is a body thrown into a uniform gravitational field of earth (near to the surface of earth surface) which follows a path determined entirely by the effects of acceleration due to gravity and air resistance. One of the conditions to study projectile motion is to neglect the effects of air resistance.
As the acceleration due to gravity is always in the direction of vertical downward, the projectile will always be confined in the vertical plane. If the projectile is thrown vertically upward the motion is one dimensional, and if the projectile is thrown inclined to the vertical, the motion is two dimensional.
The time to rise is equal to the time of fall..

Note:
In the case of vertical projectiles, the horizontal distance is calculated as $x = (u\operatorname{Cos} \theta )t$ where t is time. The equation of trajectory represents a parabola with a equation, $y = x\tan \theta - \dfrac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}$ where symbols have their usual meanings. Time of flight $T = \dfrac{{2u\operatorname{Sin} \theta }}{g}$ . Maximum height attained ${H_{\max }} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ . Range is calculated by the formula $R = \dfrac{{{u^2}\sin 2\theta }}{g}$ . The condition of maximum range is $R = \dfrac{{{u^2}\sin 2\theta }}{g} = \dfrac{{{u^2}\sin 2(90^\circ - \theta )}}{g} = \dfrac{{{u^2}}}{g}$ .