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.
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Correct Answer: Option n
Explanation:
(a) If initial velocity = U Then horizontal component
\(U_x = Ucos\theta\)
Ucos\(\theta\) = 50
U = \(\frac{50}{cos \theta} = \frac{50}{cos 30^o}\) = 57.7m/s
(b) But at maximum height V = O
V\(_y\) = O
V\(_x\) = U\(_x\) = 50m/s
(a) If initial velocity = U Then horizontal component
\(U_x = Ucos\theta\)
Ucos\(\theta\) = 50
U = \(\frac{50}{cos \theta} = \frac{50}{cos 30^o}\) = 57.7m/s
(b) But at maximum height V = O
V\(_y\) = O
V\(_x\) = U\(_x\) = 50m/s