Answer:
A. t = 0.37 s
B. Xmax = 2.46 m.
Explanation:
A. The ball's time of flight can be found as follows:
[tex] t_{f} = \frac{2v_{0y}}{g} [/tex]
Where:
t(f) is the time of flight
g is the gravity = 9.81 m/s²
[tex]v_{0y}[/tex] is the initial speed in the vertical direction (y)
[tex] t_{f} = \frac{2v_{0y}}{g} = \frac{2*7 m/s*sin(15)}{9.81 m/s^{2}} = 0.37 s [/tex]
Hence the ball's time of flight is 0.37 s.
B. To find the maximum distance we need to use the next equation:
[tex] X_{max} = X_{0} + v_{0x}t_{v} - \frac{1}{2}a_{x}t_{v}^{2} [/tex]
Where:
X₀ is the initial position = 0 (the origin in the Cartesian coordinate)
[tex]a_{x}[/tex] is the acceleration = 0.66 m/s²
[tex] X_{max} = 0 + 7 m/s*cos(15)*0.37 s - \frac{1}{2}0.66 m/s^{2}*(0.37 s)^{2} = 2.46 m [/tex]
Therefore, the horizontal distance the ball travels before it returns to the field is 2.46 m.
I hope it helps you!
