If a projectile is shot at an angle of #(2pi)/3# and at a velocity of #19 m/s#, when will it reach its maximum height?

1 Answer
Jul 1, 2017

Answer:

#t = 1.68# #"s"#

Explanation:

We're asked to find the time when a projectile reaches its maximum height, given its initial velocity.

When the projectile is at its maximum height, its instantaneous #y#-velocity is zero.

We can use the equation

#v_y = v_(0y) - g t#

to find the time.

The initial #y#-velocity #v_(0y)# is given by

#v_(0y) = v_0sinalpha_0#

#v_(0y) = (19color(white)(l)"m/s")sin((2pi)/3) = color(red)(16.5# #color(red)("m/s"#

And #g = 9.81# #"m/s"^2#

Plugging in known values, we have

#0 = color(red)(16.5)color(white)(l)color(red)("m/s") - (9.81color(white)(l)"m/s"^2)t#

#t = (color(red)(16.5)color(white)(l)color(red)("m/s"))/(9.81color(white)(l)"m/s"^2) = color(blue)(1.68# #color(blue)("s"#