# Direction of velocity

1. Jun 6, 2010

### thereddevils

1. The problem statement, all variables and given/known data

A particle is projected from the top of a vertical cliff with a speed of 40 m/s at an angle of elevation of 45 degrees towatds the sea . The particle hits the surface of the sea at a point whose horizontal distance from the cliff is 200 m . Find the direction of the velocity of the particle when it hits the surface of the sea . [Take g=10 m/s^2]

2. Relevant equations

3. The attempt at a solution

For horizontal motion ,

s=vt

200=40 cos 45 t

t=5 root (2)

For vertical motion ,

v=u+at

v=40 sin 45 -10(5 root 2)

=-(60)/(root 2)

This is the magnitude and i hv no idea how to find its direction

2. Jun 6, 2010

### Staff: Mentor

This gives you the vertical component of the velocity, not the magnitude. What's the horizontal component? Use the components to find the angle.

3. Jun 6, 2010

### thereddevils

argh , the answer is just right in front of me and i missed it . Thank you Sir .