Explanation:
It is given that,
Initial speed of the ball, u = 4.05 m/s
The roof is pitched at an angle of 40 degrees below the horizontal
Height of the edge above the ground, h = 4.9 m
Let for t time the baseball spends in the air. It can be solved using the second equation of motion as :
[tex]h=ut+\dfrac{1}{2}gt^2[/tex]
[tex]h=u\ sin\theta t+\dfrac{1}{2}gt^2[/tex]
[tex]4.9=4.05\ sin(40)t+\dfrac{1}{2}\times 9.8 t^2[/tex]
On solving the above equation, we get, t = 0.769 seconds
Let x is the horizontal distance from the roof edge to the point where the baseball lands on the ground. It can be calculated as :
[tex]x=u\ cos\theta\times t[/tex]
[tex]x=4.05\ cos(40)\times 0.769[/tex]
x = 2.38 meters
Hence, this is the required solution.
