A particle is projected at an angle of 30^\circ 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\(^{-2}\)]
Verified Correct Answer: .
Physical Principles & Numerical Solution
(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
Source: Myschool Past Questions Archive (Record #53636)