The horizontal component of the initial speed of a particle projected at 30° to the horizontal is 50 ms\(^{-1}\). If the acceleration cf free fall due to gravity is 10ms\(^{-2}\), determine its: (a) initial speed; (b) speed at maximum height reached.
The particle is projected at \(30^\circ\) to the horizontal, and its horizontal component of speed is \(u_x = 50\,\text{ms}^{-1}\).
(a) Initial speed \(u\)
The horizontal component is \(u_x = u\cos\theta\), so:
\[ u = \frac{u_x}{\cos\theta} = \frac{50}{\cos 30^\circ} = \frac{50}{0.866} = 57.7\,\text{ms}^{-1} \]
(b) Speed at maximum height
At the maximum height the vertical component of velocity is zero, so the speed there equals the (unchanged) horizontal component:
\[ v = u\cos\theta = 50\,\text{ms}^{-1} \]