Determine the time after which the velocity of a ball,projected with speed v0 at an angle a0 from a point, becomes perpendicular to the velocity of projection. The resistance of air is neglected.
At any timr t,the horizontal component x and the vertical component y of displacement of the projected ball with an angle a to horizontal and initial velocity v0 is given by:
x = v0t cosa.
y = votsina -(1/2) gt^2.
At the start the slope of the ball to horiziontal = tana....(1)
At any time t the slope of the ball = dy/dx = (dy/dt)/(dx/dt)
= (v0*t cosa)'/(v0*tsina-(1/2)t^2)'
dy/dx =(v0sina -gt)/v0cosa.........(2)
If the direction of the ball is perpendicular to the initial direction, then
dy/dx*tana = -1. Use the value of dy/dx obtained in (2):
((vosina-gt)/vocosa)* tana = -1
vosin^2 a -gt sina = -v0cos^2a
v0 (sin^2a+cos^2a ) = gt sina
v0 = gtsina.
Therfore t = v0/gsina.
So at time t = v0/gsina , the ball will be moving in the projctile perpendicular to the initial direction a.
We'll choose the time t = 0 when the ball is projected. After the time t, it's velocity vector becomes perpendicular to the velocity vector of projection v0.
We'll write the law that describes the velocity of the ball at time t:
v = v0 + g*t (1)
If v is perpendicular to v0, then the dot product of the vectors is:
v*v0 = 0
the formula for dot product is:
v*v0 = |v|*|v0|*cos (v,v0)
The angle between v and v0 is 90 degrees, so cos 90 = 0.
v*v0 = |v|*|v0|*0
v*v0 = 0
We'll substitute v by the law (1):
(v0 + g*t)*v0 = 0
We'llremove the brackets:
v0^2 + g*t*v0 = 0 (2)
We'll substitute v0 = v0*cos(pi/2 + a0) (3)
We'll substitute (3) in (2):
v0^2 + g*t*v0*cos(pi/2 + a0) = 0
But cos(pi/2 + a0) = sin a0
v0^2 + g*t*v0*sin a0 = 0
We'll factorize by v0:
v0(v0 + g*t*sin a0) = 0
If v0 is different of zero, then the other factor must be zero:
v0 + g*t*sin a0 = 0
We'll subtract v0:
g*t*sin a0 = -v0
We'll divide by g*sin a0:
t = - v0/g*sin a0
The velocity of the ball becomes perpendicular to the velocity of projection after the time t = - v0/g*sin a0.