# projectile motion

• May 19th 2009, 06:45 PM
slaypullingcat
projectile motion
A fast bowler runs in to bowl and releases the ball froma height of 2.4m above the ground at a speed of 144km/h at an angle of 7 degrees below the horizontal. Take the origin to be the point where the ball is released and g=-10m/s/s.
(a) show that the coordinates of the ball after t seconds is given by

1) x=40t cos 7 and
2) y=2.4-40t sin 7-5t^2

My question is where did the 40 come from and where has the velocity of 144km/h gone?

Thank you
• May 19th 2009, 06:57 PM
skeeter
Quote:

Originally Posted by slaypullingcat
A fast bowler runs in to bowl and releases the ball froma height of 2.4m above the ground at a speed of 144km/h at an angle of 7 degrees below the horizontal. Take the origin to be the point where the ball is released and g=-10m/s/s.
(a) show that the coordinates of the ball after t seconds is given by

1) x=40t cos 7 and
2) y=2.4-40t sin 7-5t^2

My question is where did the 40 come from and where has the velocity of 144km/h gone?

Thank you

unit conversion ...

144 km/hr = 40 m/s
• May 19th 2009, 07:00 PM
slaypullingcat
Ha, jeez i'm thick some times. Cheers.