Understanding Projectile Motion
Projectile motion is the motion of an object thrown or projected into the air, subject only to the acceleration of gravity. We analyze this by splitting the motion into two independent components: horizontal ($x$) and vertical ($y$).
Given Data
- Initial velocity ($v_0$) = $37 \, \text{m/s}$
- Launch angle ($\theta$) = $53^\circ$
- Time ($t$) = $2 \, \text{s}$
- Acceleration due to gravity ($g$) = $9.8 \, \text{m/s}^2$ (downward)
First, we resolve the initial velocity into components:
- $v_{0x} = v_0 \cos(53^\circ) \approx 37 \times 0.6 = 22.2 \, \text{m/s}$
- $v_{0y} = v_0 \sin(53^\circ) \approx 37 \times 0.8 = 29.6 \, \text{m/s}$
1. Position of the Ball
- Horizontal position: $x = v_{0x} \cdot t = 22.2 \times 2 = 44.4 \, \text{m}$
- Vertical position: $y = v_{0y} \cdot t - 0.5 \cdot g \cdot t^2 = 29.6 \times 2 - 0.5 \times 9.8 \times (2^2) = 59.2 - 19.6 = 39.6 \, \text{m}$
2. Velocity After 2 Seconds
- Horizontal velocity ($v_x$) = $v_{0x} = 22.2 \, \text{m/s}$ (remains constant)
- Vertical velocity ($v_y$) = $v_{0y} - g \cdot t = 29.6 - 9.8 \times 2 = 10.0 \, \text{m/s}$
Magnitude of velocity ($v$): $v = \sqrt{v_x^2 + v_y^2} = \sqrt{22.2^2 + 10.0^2} = \sqrt{492.84 + 100} = \sqrt{592.84} \approx 24.35 \, \text{m/s}$
Direction ($\alpha$): $\alpha = \tan^{-1}(v_y / v_x) = \tan^{-1}(10.0 / 22.2) \approx 24.25^\circ$ above the horizontal.