If a projectile is shot at an angle of #(2pi)/3# and at a velocity of #1 m/s#, when will it reach its maximum height?

1 Answer
Jul 21, 2017

Answer:

#t = 0.0883# #"s"#

Explanation:

We're asked to find the time #t# when a particle reaches its maximum height, given its initial velocity.

To do this, we can use the equation

#v_y = v_0sinalpha_0 - g t#

where

  • #v_y# is the velocity at time #t# (this will be #0#, since at a particle's maximum height its instantaneous #y#-velocity is #0#)

  • #v_0# is the initial speed (give as #1# #"m/s"#)

  • #alpha_0# is the launch angle (given as #(2pi)/3#)

  • #g# is the acceleration due to gravity near earth's surface (#9.81# #"m/s"^2#)

  • #t# is the time (what we're trying to find)

Plugging in known values, we have

#0 = (1color(white)(l)"m/s")sin((2pi)/3) - (9.81color(white)(l)"m/s"^2)t#

#t = (0.866color(white)(l)"m/s")/(9.81color(white)(l)"m/s"^2) = color(red)(0.0883# #color(red)("s"#