Question

A projectile is shot from the ground at an angle of 60 degrees with respect to the horizontal, and it lands on the ground 5 seconds later. Find:
a. the horizontal component of initial velocity
b. the vertical component of initial velocity
c. the magnitude and the direction of velocity when the projectile hits the ground

Answer 1

Answer on Question #40583 – Physics – Mechanics | Kinamatics | Dynamics

1. A projectile is shot from the ground at an angle of 60 degrees with respect to the horizontal, and it lands on the ground 5 seconds later. Find: a) the horizontal component of initial velocity; b) the vertical component of initial velocity; c) the magnitude and the direction of velocity, when the projectile hits the ground.

$$\begin{array}{l l} \alpha = 60{}^{\circ} & \text{Solution.} \\ t_{1} = 5\,s & \text{Let denote } y = v_{0} \sin \alpha \cdot t - \frac{gt^{2}}{2} \text{ as initial velocity of the projectile.} \\ v_{0x}, v_{0y}, v_{1}, \alpha_{1} - ? & \text{In the horizontal direction the movement is with the constant velocity:} \\ \end{array}$$ $$v_{x} = v_{0} \cos \alpha, \quad x = v_{0} \cos \alpha \cdot t.$$

In vertical direction the movement is with the constant acceleration $g$ :

$$v_{y} = v_{0} \sin \alpha - gt, \quad y = v_{0} \sin \alpha \cdot t - \frac{gt^{2}}{2}.$$

At the moment $t = t_1$ , the $y$ -coordinate is zero.

$$y = 0, \quad 0 = v_{0} \sin \alpha \cdot t_{1} - \frac{gt_{1}^{2}}{2}, \quad v_{0} = \frac{gt_{1}}{2 \sin \alpha}.$$

So, the horizontal and vertical components of the initial velocity are:

$$v_{0x} = \frac{gt_{1}}{2} ctg\alpha, \quad v_{0y} = \frac{gt_{1}}{2}.$$

The magnitude and the direction of the final velocity are:

$$v_{1} = \sqrt{v_{1x}^{2} + v_{1y}^{2}} = \sqrt{\left(\frac{gt_{1}}{2} ctg\alpha\right)^{2} + \left(\frac{gt_{1}}{2} - g \cdot t_{1}\right)^{2}}, \quad v_{1} = \frac{gt_{1}}{2 \sin \alpha},$$ $$\alpha_{1} = arctg \frac{v_{1y}}{v_{1x}} = arctg \frac{\frac{gt_{1}}{2} - g \cdot t_{1}}{\frac{gt_{1}}{2} ctg\alpha}, \quad \alpha_{1} = -\alpha.$$

Let check the dimensions.

$$\left[ v_{0x} \right] = \left[ v_{0y} \right] = \left[ v_{1} \right] = \frac{m}{s^{2}} \cdot s = \frac{m}{s}, \quad \left[ \alpha_{1} \right] = 0.$$

Let evaluate the quantities.

$$v_{0x} = \frac{9.8 \cdot 5}{2} \cdot ctg \cdot 60{}^{\circ} = 14.1 \left(\frac{m}{s}\right), \quad v_{0y} = \frac{9.8 \cdot 5}{2} = 24.5 \left(\frac{m}{s}\right), \quad v_{1} = \frac{9.8 \cdot 5}{2 \cdot \sin 60{}^{\circ}} = 28.3 \left(\frac{m}{s}\right).$$

Answer: a) $14.1 \frac{m}{s}$ ; b) $24.5 \frac{m}{s}$ ; c) $28.3 \frac{m}{s}, -60{}^{\circ}$ .