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
While ignoring air friction...
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
am sorry but could you explain how you got 7.7 revolutions as your final answer ? I do 19sin of 55 I get =15.5639 and then Im lost , thanks again
The 15.5639 is the time in second while the 7.7 is the rev per second. So the final answer should be the time times revolutions per second.
To find the number of revolutions the ball has made in the air, we need to determine the total time the ball is in the air and then calculate the number of revolutions based on the spin rate.
First, let's calculate the time the ball is in the air. We can use the given linear speed and the angle at which it was thrown.
The horizontal component of the linear speed is given by:
Vx = V * cos(angle)
Vx = 19 m/s * cos(55°)
Vx ≈ 19 m/s * 0.574
Vx ≈ 10.94 m/s
Now, we can calculate the time the ball is in the air using the horizontal component of the linear speed and the distance traveled horizontally.
Time = Distance / Speed
Time = 0 (since the starting and ending height is the same) / Vx
Time = 0 / 10.94 m/s
Time = 0 seconds
As we find that the time is 0 seconds, it implies that the ball hasn't traveled horizontally and will fall straight down. Therefore, the ball hasn't made any revolutions in the air.
So, the ball has not made any revolutions in the air.