A second baseman tosses the ball to the first baseman, who catches it at the same level from which it was thrown. The throw is made with an initial speed of 19.0m/s at an angle of 33.0 ∘ above the horizontal. How long is the ball in the air?
initial speed up Vi = 19 sin 33 = 10.35m/s
v = Vi - 9.81 t
at top
0 = 10.35 - 9.81 t
so t = 1.055 seconds to TOP
so
2*1.05 = 2.11 seconds total in air
To find out how long the ball is in the air, we need to consider the horizontal and vertical components of the motion separately.
First, let's find the time it takes for the ball to reach its highest point. We can use the vertical component of the initial velocity and the acceleration due to gravity (9.8 m/s^2) to find the time of flight to the highest point.
Using the equation: v = u + at, where:
- v is the final velocity (0 m/s at the highest point)
- u is the initial vertical velocity (19.0 m/s * sin(33.0°))
- a is the acceleration due to gravity (-9.8 m/s^2)
- t is the time taken to reach the highest point
0 = (19.0 m/s * sin(33.0°)) - (9.8 m/s^2 * t)
Simplifying the equation, we have:
19.0 m/s * sin(33.0°) = 9.8 m/s^2 * t
Now, let's solve for t:
t = (19.0 m/s * sin(33.0°)) / (9.8 m/s^2)
Next, let's find the total time of flight, which is twice the time taken to reach the highest point since the ball reaches the same horizontal level from which it was thrown.
Total time of flight = 2 * t
Now, substitute the value of t into the equation to find the total time of flight.
Finally, calculate the total time of flight to get the answer.