Question

1,a cricket ball of mass m is hitted at an angle 45 degree to the horizontal with velocity v.what is its kinetic energy at the topmost point?

Answer 1

A cricket ball of mass $m$ is hitted at an angle 45 degree to the horizontal with velocity $v$. What is its kinetic energy at the topmost point?

Solution:

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

$$V_y \neq \text{const}, V_x = \text{const} = V \cos \alpha$$ $$\alpha = 45{}^\circ, \cos 45{}^\circ = \frac{1}{\sqrt{2}}$$ $$V_x = V \cos \alpha = \frac{V}{\sqrt{2}}$$

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:

$$2: V_2 = V_x = \frac{V}{\sqrt{2}} \quad (1)$$

Formula of the kinetic energy:

$$E_k = \frac{m V_2^2}{2} \quad (2)$$ $$(1)\text{in} \quad (2): E_k = \frac{m V_2^2}{2} = \frac{m V^2}{4} = 0.25 m V^2$$

Answer: $E_k = 0.25 m V^2$