Question

A particle is projected with speed V at an angel ∝ to the horizontal. Find the speed of the particle when it is at a height h.

Answer 1

Answer on Question #49707, Physics, Mechanics | Kinematics | Dynamics

A particle is projected with speed $V$ at an angel $\propto$ to the horizontal. Find the speed of the particle when it is at a height $h$ .

Solution:

Projectile motion is a form of motion in which an object or particle (called a projectile) is thrown near the earth's surface, and it moves along a curved path under the action of gravity only.

In projectile motion, the horizontal motion and the vertical motion are independent of each other; that is, neither motion affects the other.

The horizontal component of the velocity of the object remains unchanged throughout the motion. The vertical component of the velocity increases linearly, because the acceleration due to gravity is constant $(g = 9.81\mathrm{m} / \mathrm{s}^2)$ .

$$v _ {x} = v _ {0} \cos \alpha$$ $$v _ {0 y} = v _ {0} \sin \alpha$$

Equations related to trajectory motion are given by

Horizontal distance, $x_{max} = v_{0x}t$

Vertical distance, $y = h = \frac{v_{0y}^2 - v_y^2}{2g}$

Thus,

$$v _ {y} ^ {2} = v _ {0 y} ^ {2} - 2 g h$$

The speed of the particle when it is at a height $h$ is

$$\begin{array}{l} v = \sqrt {v _ {x} ^ {2} + v _ {y} ^ {2}} = \sqrt {(v _ {0} \cos \alpha) ^ {2} + (v _ {0} \sin \alpha) ^ {2} - 2 g h} = \sqrt {v _ {0} ^ {2} (\cos^ {2} \alpha + \sin^ {2} \alpha) - 2 g h} = \\ = \sqrt {v _ {0} ^ {2} - 2 g h} \\ \end{array}$$

Answer: $v = \sqrt{v_0^2 - 2gh}$

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