Question

A particle is projected at an angle of 60 to the horizontal with a speed of 20ms calculate the total time of flight of the particle and speed of the particle at its maximum height (g=10m/s)

Answer 1

Answer:

Explanation:

Total time of flight is given by 2usin60/g = 2×20×1.73/2×10 = 3.46seconds

Velocity at maximum height is always zero for vertical component and remains same for horizontal component

Thus V = ucos60 = 20 × 1 /2 = 10m/s