Question

A stone thrown from the surface of the earth at a speed of 10m / s at an angle of 600 to the horizontal. The radius of curvature of the upper root point of the trajectory

Answer 1

A stone thrown from the surface of the earth at a speed of $10\mathrm{m} / \mathrm{s}$ at an angle of 600 to the horizontal. The radius of curvature of the upper root point of the trajectory

Solution:

Gravitational acceleration change velocity amount only on the vertical component, the horizontal component then remains constant:

$V_{x} = const = V\cos \alpha$ (from a right triangle)

$\alpha = 60{}^{\circ}, \cos 60{}^{\circ} = \frac{1}{2}$

$V_{x} = V\cos \alpha = \frac{V}{2}$

At the topmost point of the trajectory velocity of the body is equal to the horizontal component of the initial velocity, so that the vertical component of the velocity is zero:

2: $V_{2} = V_{x} = \frac{V}{2}$

Formula for the centripetal acceleration at the topmost point (2):

$a_{c} = \frac{V_{x}^{2}}{R} = g$ (the only acceleration that acts on the ball)

$R = \frac{V_x^2}{a_c} = \frac{\left(\frac{V}{2}\right)^2}{g} = \frac{V^2}{4g} = \frac{\left(10\frac{m}{s}\right)^2}{4\cdot 9.8\frac{m}{s^2}} = 2.6m$

Answer: the radius of curvature of the upper root point of the trajectory is $2.6m$