A quarterback threw a football at 27 m/s at a certain angle above the horizontal. If it took the ball 3.3 s to reach the top of its path, how long was it in the air?
s
6.6 seconds....
To find how long the ball was in the air, we can break down the motion into horizontal and vertical components.
First, let's find the time it takes for the ball to reach its highest point. Since the ball reaches its highest point at the top of its path, the vertical component of its velocity at this point is 0 m/s.
We can use the following equation to find the time it takes to reach the top of the path:
v_y = v_0y + a_y * t
Where:
v_y is the vertical component of velocity at the top of the path (0 m/s),
v_0y is the initial vertical component of velocity (unknown),
a_y is the acceleration due to gravity (-9.8 m/s^2),
and t is the time it takes to reach the top of the path (3.3 s).
0 = v_0y + (-9.8) * 3.3
Now, solve for v_0y:
v_0y = 9.8 * 3.3
Next, let's find the total time the ball is in the air. Since the ball follows a symmetrical path, the total time in the air is twice the time it takes to reach the top of the path.
Total time in the air = 2 * 3.3 s
Therefore, the ball was in the air for a total of 6.6 seconds.