Answer:
a) [tex] v_{0y} = 9.72 m/s [/tex]
b) [tex] v_{0x} = 20.85 m/s [/tex]
c) [tex] t = 2.01 s [/tex]
d) [tex] h_{max} = 4.82 m [/tex]
Explanation:
a) The initial vertical velocity is given by:
[tex] v_{0y} = v*sin(\theta) [/tex]
Where:
θ: 25°
v: is the magnitude of the speed = 23 m/s
[tex] v_{0y} = 23 m/s*sin(25) = 9.72 m/s [/tex]
b) The initial horizontal velocity can be calculated as follows:
[tex] v_{0x} = v*cos(\theta) = 23 m/s*cos(25) = 20.85 m/s [/tex]
c) The flight time can be calculated using the following equation:
[tex] v_{0x} = \frac{x}{t} [/tex]
Where:
x: is the total distance = 42 m
[tex] t = \frac{x}{v_{0x}} = \frac{42 m}{20.85 m/s} = 2.01 s [/tex]
d) The maximum height is given by:
[tex] v_{fy}^{2} = v_{0y}^{2} - 2gh_{max} [/tex]
Where:
[tex] v_{fy}[/tex]: is the final vertical velocity =0 (at the maximum heigth)
g: is the gravity = 9.81 m/s²
[tex]h_{max} = \frac{v_{0y}^{2}}{2g} = \frac{(9.72 m/s)^{2}}{2*9.81 m/s^{2}} = 4.82 m[/tex]
I hope it helps you!
