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\(^{-2}\)]
(a) total time of flight of the particle;
(b) speed of the particle at its maximum height. [ g = 10 ms\(^{-2}\)]
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Correct Answer: Option n
Explanation:
(a) T = \(\frac{2Usin\theta}{g} = \frac{2 \times 250 \times sin 30}{10}\)
= 25s
(b) Speed at h\(_{\text{max}} Ucos \theta\)
= 250 cos 30\(^o\) = 216m/s
(a) T = \(\frac{2Usin\theta}{g} = \frac{2 \times 250 \times sin 30}{10}\)
= 25s
(b) Speed at h\(_{\text{max}} Ucos \theta\)
= 250 cos 30\(^o\) = 216m/s