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

Answer:

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)