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