Answer:
A) t = 0.55 s
B) x = 24.8 m
Explanation:
A) We can find the time at which the ball will be in the air using the following equation:
[tex] y_{f} = y_{0} + v_{0y}t - \frac{1}{2}gt^{2} [/tex]
Where:
[tex]y_{f}[/tex] is the final height= 0
[tex]y_{0}[/tex] is the initial height= 1.5 m
[tex]v_{0y}[/tex] is the component of the initial speed in the vertical direction = 0 m/s
t: is the time =?
g: is the gravity = 9.81 m/s²
[tex] 0 = 1.5 m - \frac{1}{2}9.81 m/s^{2}t^{2} [/tex]
By solving the above equation for t we have:
[tex] t = \sqrt{\frac{2*1.5 m}{9.81 m/s^{2}}} = 0.55 s [/tex]
Hence, the ball will stay 0.55 seconds in the air.
B) We can find the distance traveled by the ball as follows:
[tex] x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2} [/tex]
Where:
a: is the acceleration in the horizontal direction = 0
[tex]x_{f}[/tex] is the final position =?
[tex]x_{0}[/tex] is the initial position = 0
[tex]v_{0x}[/tex] is the component of the initial speed in the horizontal direction = 45 m/s
[tex] x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2} [/tex]
[tex] x_{f} = 0 + 45 m/s*0.55 s + 0 = 24.8 m [/tex]
Therefore, the ball will travel 24.8 meters.
I hope it helps you!
