If a projectile is shot at an angle of (2pi)/3 and at a velocity of 5 m/s, when will it reach its maximum height?

1 Answer
Feb 22, 2018

time taken to reach maximum height = 0.4 s (approx)

Explanation:

The velocity vector of the projectile has a x- component and y - component .
since we are dealing with height , we just need the y-component .
velocity in y direction = 5 sin ((2pi)/3) ms^-1
=(5sqrt 3) /2 ms^-1

now using initial velocity u = (5 sqrt3 )/2 ms^-1 ,

final velocity v = 0 ms^-1, acceleration due to gravity a = -9.8 ms^-2 .

we have v = u + at

=> 0 = (5sqrt3) /2 + -9.8 t

=> 9.8 t = (5sqrt3) /2

=> t = (5sqrt3) /(2×9.8)

=> t= 0.4 s (approx)