Answer
64.8k+ views
Hint: Kicked football forms a projected motion, therefore we define horizontal range;
Horizontal range of a projectile is the distance from the point of projection to the point where the projectile comes back to the plane of projection.
Formula for horizontal range is:
$R = \dfrac{{{v^2}\sin 2\theta }}{g}$ (v is the velocity, g is the gravitational acceleration, $\theta $ is the angle at which the object is projected) which is directly proportional to the velocity of the object being projected.
Complete step by step solution:
Let us define Horizontal range and horizontal component of velocity in detail.
It is the total horizontal distance from the point of projection to the point where the projectile comes back to the plane of projection. It is denoted by R;
In order to calculate horizontal range R, we shall consider horizontal motion of the projectile. The horizontal motion is uniform. It takes place with constant velocity of horizontal component $v\cos \theta $.
$R = v\cos \theta \times $ time of flight
where time of flight is given by;
$T = \dfrac{{2u\sin \theta }}{g}$
Therefore horizontal range is given by:
$R = v\cos \theta \times \dfrac{{2v\sin \theta }}{g}$.....................1
Component of horizontal velocity: Horizontal component of the velocity is the component of the velocity at which the velocity makes angle of projection it is given as;
$v\cos \theta $, which is directly proportional to horizontal range.
In the figure of the question, the fourth path of the kicked football has the maximum range, thus the fourth path of the football has the highest component of the horizontal velocity, then third, second and first.
Note: Examples of the objects of which shows the projectile motion are: a bomb released from a level flight, a bullet fired from a gun, an arrow released from bow, a javelin thrown by athlete. In all these motions we must neglect the resistance made by air and rotation of earth and the effect due to curvature of earth.
Horizontal range of a projectile is the distance from the point of projection to the point where the projectile comes back to the plane of projection.
Formula for horizontal range is:
$R = \dfrac{{{v^2}\sin 2\theta }}{g}$ (v is the velocity, g is the gravitational acceleration, $\theta $ is the angle at which the object is projected) which is directly proportional to the velocity of the object being projected.
Complete step by step solution:
Let us define Horizontal range and horizontal component of velocity in detail.
It is the total horizontal distance from the point of projection to the point where the projectile comes back to the plane of projection. It is denoted by R;
In order to calculate horizontal range R, we shall consider horizontal motion of the projectile. The horizontal motion is uniform. It takes place with constant velocity of horizontal component $v\cos \theta $.
$R = v\cos \theta \times $ time of flight
where time of flight is given by;
$T = \dfrac{{2u\sin \theta }}{g}$
Therefore horizontal range is given by:
$R = v\cos \theta \times \dfrac{{2v\sin \theta }}{g}$.....................1
Component of horizontal velocity: Horizontal component of the velocity is the component of the velocity at which the velocity makes angle of projection it is given as;
$v\cos \theta $, which is directly proportional to horizontal range.
In the figure of the question, the fourth path of the kicked football has the maximum range, thus the fourth path of the football has the highest component of the horizontal velocity, then third, second and first.
Note: Examples of the objects of which shows the projectile motion are: a bomb released from a level flight, a bullet fired from a gun, an arrow released from bow, a javelin thrown by athlete. In all these motions we must neglect the resistance made by air and rotation of earth and the effect due to curvature of earth.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)