A baseball is thrown at an angle of 26◦ relative
to the ground at a speed of 25 m/s. The ball
is caught 50.2046 m from the thrower.
The acceleration due to gravity is
9.81 m/s2 .
How long is it in the air?
Answer in units of s.
To find out how long the baseball is in the air, we can use the horizontal motion and vertical motion of the ball separately.
Let's start with the vertical motion of the ball.
The initial vertical velocity (Vi) of the ball can be calculated using the initial speed (25 m/s) and the angle of projection (26 degrees) using the formula:
Vi = initial speed * sin(angle)
So, Vi = 25 m/s * sin(26 degrees)
Next, we need to find the time it takes for the ball to reach its maximum height. The time taken to reach the maximum height is half of the total time the ball spends in the air, as the time taken to reach the highest point is the same as the time taken to descend from the highest point to the ground.
The time taken to reach the maximum height (t_max) can be calculated using the formula:
t_max = (final vertical velocity - initial vertical velocity) / acceleration due to gravity
At the maximum height, the vertical velocity becomes 0 because the ball momentarily stops before falling back down. The final vertical velocity (Vf) is 0, acceleration due to gravity (g) is 9.81 m/s^2.
Therefore, t_max = (0 - Vi) / g
Now, let's find the total time the ball is in the air by calculating the time it takes for the ball to reach the maximum height and then double that time.
Total time in the air = 2 * t_max
Finally, you can substitute the values and calculate the time the baseball is in the air.