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M1 projectiles, quick question

untitled.JPG

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?
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.
Reply 2
Avatar for xtal
xtal
OP
Is this correct?

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

untitled.JPG

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

I'm still not sure what it means by 'below the horizontal'.
(edited 12 years ago)
Reply 3
Avatar for xtal
xtal
OP
Anyone?
Reply 4
Avatar for TenOfThem
TenOfThem
18
Well the horizontal is the horizontal line and your angle is below that
Reply 5
Avatar for Sarabande
Sarabande
15
Isn't projectile motion M2?
Reply 6
Avatar for TenOfThem
TenOfThem
18
Original post by Sarabande
Isn't projectile motion M2?


I guess it must be board dependant

It is M1 for AQA
Reply 7
Avatar for Sarabande
Sarabande
15
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.

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