Question

The following equation describes the horizontal range of a projectile, fired from a height of zero, at an angle of ?θ, with an initial velocity of v:

?= ?^2 sin(2?) / ?,

where g is the acceleration of gravity. Use kinematics to derive this equation by assuming the magnitude of the initial velocity to be v. Hint: the horizontal and vertical components of the initial velocity can be expressed as ??=????(?)vx=vcos(θ) and ??=????(?)vy=vsin(θ). Another hint: you aren't given any numbers but you can still proceed as usual (i.e., find an expression for t using kinematics in the y direction and then plug it into a kinematic equation in the x direction). A third hint: the double angle identity states: sin(2?)=2sin(?)cos(?)

Answer 1

Solution: We have,

Acceleration of the projectilie along y-axis, and along x -axis, .

If a projectile is fired with magnitude of initial velocity at an angle of with horizontal. Then x- and y- components are:

     

As we know that at heighest point, vertical velocity of the projectile is zero i.e., .

Now, using equation of kinematics along vertical direction we can write as:

     ......(1)

So, the time of fight is given by:

  

Since, there is no acceleration of the projectile in x direction so, velocity of the projectile remains same throughout the motion. So, using kinematics we can write as:  

  

Where is the horizontal range of a projectile