A particle is projected at an angle of 30° to the horizontal with a speed of 250 ms\(^{-1}\). Calculate the; (a) total time of flight of the particle; (b) s...
Question 1 Report
A particle is projected at an angle of 30° to the horizontal with a speed of 250 ms\(^{-1}\). Calculate the;
(a) total time of flight of the particle;
(b) speed of the particle at its maximum height. [ g = 10 ms\(^{-2}\)]
Answer Details
Given \(u=250\,\text{ms}^{-1}\), \(\theta=30^{\circ}\), \(g=10\,\text{ms}^{-2}\).
(a) Total time of flight.
\[ T=\frac{2u\sin\theta}{g}=\frac{2\times250\times\sin30^{\circ}}{10}=\frac{2\times250\times0.5}{10}=\frac{250}{10}=25\,\text{s} \](b) Speed at maximum height. At the highest point the vertical component of velocity is zero, so the speed there is just the (unchanged) horizontal component:
\[ v=u\cos\theta=250\times\cos30^{\circ}=250\times0.866=216.5\,\text{ms}^{-1} \]Download The App On Google Playstore
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