Different results for the Range of a projectile

[RIJU101]

TL;DR Summary

My question is from the Range of a projectile. Range is displacement of the projectile through X-axis. We all know the formula of range is R=(V²Sin2x)/g. Where "x" is the angle at which the object is thrown. When we throw the object at 0° angle the value of range will be Zero(0) that means this value concludes that object will not displace. But actually it will displace horizontally so there should be a value of Range as per the definition. Thats my question why the concept and result differs?

May be It's a concept problem.

 

[kuruman]

The formula for the range that you quoted is valid only if the projectile lands at the same height from which it was launched. If it is launched horizontally, it must start at some height above ground and will not return to the same height from which it was launched. If it's launched at ground level, it is already there, so zero horizontal displacement.

On edit
To complete the picture, the general expression for the range of a projectile starting with initial velocity $\mathbf v=(v_{0x},v_{0y})$ is given by
$$R=\frac{v_{0x}}{g}\left[v_{0y}+\sqrt{v_{0y}^2-2 g\Delta y}\right]$$ where $\Delta y$ is the vertical displacement, positive if the projectile lands higher and negative if it lands lower than launched. When $\Delta y =0$ the expression above reduces to the one you quoted.

 

[RIJU101]

The formula for the range that you quoted is valid only if the projectile lands at the same height from which it was launched. If it is launched horizontally, it must start at some height above ground and will not return to the same height from which it was launched. If it's launched at ground level, it is already there, so zero horizontal displacement.

On edit
To complete the picture, the general expression for the range of a projectile starting with initial velocity $\mathbf v=(v_{0x},v_{0y})$ is given by
$$R=\frac{v_{0x}}{g}\left[v_{0y}+\sqrt{v_{0y}^2-2 g\Delta y}\right]$$ where $\Delta y$ is the vertical displacement, positive if the projectile lands higher and negative if it lands lower than launched. When $\Delta y =0$ the expression above reduces to the one you quoted.

Yes you are right. Sorry I didn't explained it correctly. I'm not talking about throwing the object from a certain height above the ground, in that case the object will displace a bit and vertically fall on the ground then range will be zero. But I'm considering a case where I'm standing on a ground and project a ball(not from a height, very close to ground) in 0°projection angle that means the ball will go horizontally, it will roll on the ground. In that case how the range can be zero.

 

[RIJU101]

The formula for the range that you quoted is valid only if the projectile lands at the same height from which it was launched. If it is launched horizontally, it must start at some height above ground and will not return to the same height from which it was launched. If it's launched at ground level, it is already there, so zero horizontal displacement.

On edit
To complete the picture, the general expression for the range of a projectile starting with initial velocity $\mathbf v=(v_{0x},v_{0y})$ is given by
$$R=\frac{v_{0x}}{g}\left[v_{0y}+\sqrt{v_{0y}^2-2 g\Delta y}\right]$$ where $\Delta y$ is the vertical displacement, positive if the projectile lands higher and negative if it lands lower than launched. When $\Delta y =0$ the expression above reduces to the one

Do you tell me how this formula derived ?

 

[A.T.]

...it will roll on the ground. In that case how the range can be zero...

This is a formula for throw distance, not for rolling distance.

 

[kuruman]

Do you tell me how this formula derived ?

Sure. The traditionall approach is to start with the kinematic equations in the x and y directions.
$x=v_{0x}t~;~~y=y_0+v_{0y}t-\frac{1}{2}gt^2$.
Let $t_{\!f}=~$the time the projectile is in the air (time of flight). At that time the projectile is at horizontal position $x_f=R$ and is vertically displaced by $\Delta y.$

From the first equation, $t_{\!f}=\dfrac{R}{v_{0x}}.$ If this is substituted in the second equation, one gets $$\Delta y=y_{\!f}-y_0=v_{0y}\frac{R}{v_{0x}}-\frac{1}{2}g\left(\dfrac{R}{v_{0x}}\right)^2.$$This is a quadratic equation that you can solve for $R$ the usual way. Only one of the two solutions gives positive $R$ values for any value of $\Delta y.$