a particle is placed at a point O on the edge of a horizontal table at height 2a above a horizontal floor. the point on the floor vertically below O is A. the particle is projected horizontally away from the table and lands on the floor at point B, where AB=a. show that the speed of projection is (1/2)(ga)^1/2. find the magnitude and direction of the velocity of the particle at the instant when it is vertically above the mid-point of AB. at time t after the projection the angle PAB is θ. show that tan θ= (1/(ga)^1/2)((4a/t)-gt)

Question

a particle is placed at a point O on the edge of a horizontal table at height 2a above a horizontal floor. the point on the floor vertically below O is A. the particle is projected horizontally away from the table and lands on the floor at point B, where AB=a. show that the speed of projection is (1/2)(ga)^1/2. find the magnitude and direction of the velocity of the particle at the instant when it is vertically above the mid-point of AB. at time t after the projection the angle PAB is  θ. show that tan θ= (1/(ga)^1/2)((4a/t)-gt)

anonymous · Answer

vertical
2a = 1/2 *gt^2
t=sqrt(4a/g)

horizontal  
a = V*t
V=a/t = a/(sqrt(4a/g) = 1/2 * sqrt(ag)

anonymous · Answer

now that you have initial horizontal velocity, try the second part...

horiontal
 a/2 =  1/2 * sqrt(ag)*t
t = sqrt(a)/sqrt(g)

vertical
Vf = g*t
Vf = g* sqrt(a)/sqrt(g) = sqrt(ag)

anonymous · Answer

find the magnitude of velocity using those two components...

anonymous · Answer

ok i will

anonymous · Answer

let me now if you're having trouble with the last bit, the angle...