Asked by Christopher

The ball spins at 7.7 rev/s. In addition, the ball is through with a linear speed of 19 m/s at an angle of 55 degrees with respect to the ground. If the ball is caught at the same height at which it left the quarterback's hand, how many revolutions has the ball made in the air?
Thanks for your help



bobpursley earlier wrote this but I still do not know where all the letters stand for. Could someone help me solve it. Thanks bobpusrley for your help

The vertical speed of the ball is 19sin55.

hf=ho+viy*t- 1/2 g t^2
0=19sin55*t-4.9t^2
solve for time in air, t.

revolutions= 7.7*timeinair

Answered by bobpursley
final height= initial height+ velocity vertical initial*time minus 1/2 acceleartion due to gravity*time^2

Answered by Christopher
but how do I know the final height, initial height, and velocity vertical initial
Answered by drwls
final height and initial height are the same, so they cancel out of BobPurley's equation. You don't have to know them.

The vertical initial velocity is 19 sin55. You calculate that.

Then solve for t as indicated by Bob
Answered by shawn
25