Projectile goes long mile!

Classical Mechanics Level 3

Let $\color{#3D99F6}{H}$ be the maximum attained by the ball when it is thrown vertically up with some velocity $\color{#3D99F6}{u}$ .

Let $\color{#3D99F6}{h}$ be the maximum height when the same ball is thrown with same velocity $\color{#3D99F6}{u}$ at an angle $\theta$ with horizontal. Let the radius of curvature at highest point be $\color{#20A900}{R}$ .

If $\color{#3D99F6}{H = 100m}$ and $\color{#3D99F6}{h = 25m}$ , Find the value of $\color{#20A900}{R}$ .

Details and Assumptions

Neglect $\color{#D61F06}{\text{Air Resistance}}$ .

The answer is 150.

1 solution

Aryan Sanghi
May 25, 2018

We know that

$\text{Radius of curvature at highest point in projectile } = \frac{u²cos²\theta}{g}$

$cos²\theta=1 - sin²\theta$

$R = \frac{u²cos²\theta}{g} = \frac{u²}{g} - \frac{u²sin²\theta}{g}$

$\frac{u²}{2g} = H$

$\frac{u²sin²\theta}{2g} = h$

$\boxed{R = \frac{u²}{g} - \frac{u²sin²\theta}{g} = 2H-2h}$

$\text{As }H=100m \text{ and } h=25m$

$R=2×100-2×25$

$\color{#3D99F6}{\boxed{R=150m}}$

The formula for radius of curvature is not given in my book, thanks! Now I will write it somewhere.

Vinayak Srivastava - 10 months, 3 weeks ago

Actually I derived it myself, so it won't be given. :P

Aryan Sanghi - 10 months, 3 weeks ago

Oh, I'll try myself then. If I am unable, I'll ask you.

Vinayak Srivastava - 10 months, 3 weeks ago

@Aryan Sanghi , I actually still could not understand what "radius" of curvature of a "parabola" means. I just need the definition, nothing in the proof, I want to prove myself.

Vinayak Srivastava - 10 months, 2 weeks ago

@Vinayak Srivastava – It is just the radius of the circle of the infinitesimally small path. Here it is Radius of curvature of trajectory

Aryan Sanghi - 10 months, 2 weeks ago

@Aryan Sanghi – Like this?

Vinayak Srivastava - 10 months, 2 weeks ago

Projectile goes long mile!

The answer is 150.

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