Question

A particle is projected from level ground in such a way that its horizontal and vertical components are 20 m/s and 10 m/s respectively. Find,
1) the maximum height of the particle
2) its horizontal distance from the point of projection when it returns to the ground.
3) the magnitude and direction of its velocity on landing

Answer 1

Answer on Question#49946 - Physics - Mechanics | Kinematics | Dynamics

1. $y\left(\frac{V_{y0}}{g}\right) = \frac{V_{y0}^2}{2g} = 5$

2. $x\left(\frac{2V_{y0}}{g}\right) = V_{x0} \frac{2V_{y0}}{g} = 40$

3. $V = 10\sqrt{5}$ . Direction is reflection from the x-axis of initial velocity.

Solution

Equations of motion:

$$x (t) = V _ {x 0} t + x _ {0}$$ $$y (t) = - \frac {g t ^ {2}}{2} + V _ {y 0} t + y _ {0}$$

Initial conditions:

$$x _ {0} = 0$$ $$y _ {0} = 0$$ $$V _ {x 0} = 2 0$$ $$V _ {y 0} = 1 0$$

Thus,

$$x (t) = V _ {x 0} t$$ $$y (t) = - \frac {g t ^ {2}}{2} + V _ {y 0} t$$

1) the maximum height of the particle

It's value of $y(t)$ at time, when $\frac{dy(t)}{dt} = 0$.

$$\frac{dy(y)}{dt} = -gt + V_{y0} = 0$$ $$t = \frac{V_{y0}}{g}$$ $$y \left(\frac{V_{y0}}{g}\right) = - \frac{g \left(\frac{V_{y0}}{g}\right)^2}{2} + V_{y0} \left(\frac{V_{y0}}{g}\right) = \frac{V_{y0}^2}{2g}$$

Substitute $V_{y0} = 10$; $g = 10$.

$$y \left(\frac{V_{y0}}{g}\right) = y(1) = \frac{10^2}{2 \times 10} = 5$$

2) the horizontal distance from the point of projection when it returns to the ground.

"Returns to ground" $\equiv y(t) = 0, t \neq 0$.

$$y(t) = - \frac{g t^2}{2} + V_{y0} t = t \left(- \frac{g t}{2} + V_{y0}\right) = 0$$ $$t = \frac{2 V_{y0}}{g}$$ $$x \left(\frac{2 V_{y0}}{g}\right) = V_{x0} \frac{2 V_{y0}}{g}$$

Substitute $V_{x0} = 20$; $V_{y0} = 10$; $g = 10$.

$$x \left(\frac{2 V_{y0}}{g}\right) = x(2) = 20 \times 2 = 40$$

3) the magnitude and direction of its velocity on landing.

$$\frac{dy}{dt} \left(\frac{2 V_{y0}}{g}\right) = -g \left(\frac{2 V_{y0}}{g}\right) + V_{y0} = - V_{y0}$$ $$\frac{dx}{dt} \left(\frac{2 V_{y0}}{g}\right) = V_{x0}$$

Hence, magnitude still the same

$$V = \sqrt{V_{x0}^2 + V_{y0}^2} = \sqrt{20^2 + 10^2} = \sqrt{400 + 100} = \sqrt{500} = 10\sqrt{5}$$

Direction is reflection from the x-axis of initial velocity.

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