contestada

Consider the video you just watched. Suppose we replace the original launcher with one that fires the ball upward at twice the speed. We make no other changes. How far behind the cart will the ball land

Respuesta :

As we know that range of the projectile motion is defined as

[tex]Range = horizontal speed \times time[/tex]

now we know that

horizontal speed = [tex]v cos\theta[/tex]

also in order to find the time of motion we will have

[tex]time = \frac{2v_y}{g}[/tex]

[tex]time = \frac{2vsin\theta}{g}[/tex]

now from the above formula of range we know that

[tex]Range = (vcos\theta)(\frac{2vsin\theta}{g}[/tex]

[tex]Range = \frac{v^2sin(2\theta)}{g}[/tex]

now from above formula we can say that if speed of the ball is twice that of initial speed then the range will increased by 4 times