[xtal]

Part e) State the magnitude and direction of the velocity of the ball as it hits the surface.

I've worked out the magnitude (20 ms^-1) but I don't understand how to work out the direction. What does '50 degrees below the horizontal' mean?

[Eduardo4tj10]

If you have the magnitude of the horizontal and vertical components of velocity of the ball as it hits the surface, then you can form a triangle and use simple trig to find the angle theta below the horizontal.

[xtal]

Is this correct?

Vertical velocity: y=20sin50 -9.8*3.13 = -15.4 ms^-1
Horizontal velocity x = 20cos50 = 12.9 ms^-1



a = tan^-1 (15.4/12.9) = 50degrees

I'm still not sure what it means by 'below the horizontal'.

[xtal]

Anyone?

[TenOfThem]

Well the horizontal is the horizontal line and your angle is below that

[Sarabande]

Isn't projectile motion M2?

[TenOfThem]

(Original post by Sarabande)
Isn't projectile motion M2?

I guess it must be board dependant

It is M1 for AQA

[Sarabande]

(Original post by TenOfThem)
I guess it must be board dependant

It is M1 for AQA

Ah. For Edexcel it's the first kinematics topic of M2.