An object is thrown with a speed u at an angle of 60 degree to the horizontal. the time elapsed between the instants when its velocity vector makes 30degree with the horizontal is?

1 Answer
Jul 7, 2017

#t = u/(gsqrt3)#

Explanation:

We're asked to find the time when the velocity vector makes an angle of #30^"o"#, given that it starts at an angle of #60^"o"#.

The initial velocity components are

#v_(0x) = ucos60^"o" = u/2# (this value doesn't change throughout motion)

#v_(0y) = usin60^"o" = (usqrt3)/2#

When the velocity vector is at an angle of #30^"o"# from the horizontal, the horizontal component is still #u/2#, but we need to find the vertical component, using trigonometry.

#tantheta = (v_y)/(v_x)#

#v_y = v_xtan30^"o" = u/2((1)/sqrt3)#

#v_y = color(red)((u)/(2sqrt3)#

Now that we know the #y#-velocity at this time, we can use the equation

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

to find the time #t# when this occurs. Plugging in known values, we have

#color(red)((u)/(2sqrt3)) = (usqrt3)/2 - g t#

#t = (color(red)((u)/(2sqrt3)) - (usqrt3)/2)/(-g)#

#t = (-(usqrt3)/3)/(-g)#

#color(blue)(t = (usqrt3)/(3g)#

or

#color(blue)(t = u/(gsqrt3))#

which is approximately

#color(blue)(t ~~ 0.0589u#