Answer:
The initial velocity of the ball is, u = 20 m/s
Hence, the amount of the time in air, [tex]t_{f}[/tex] = 2.45 s
Hence, the maximum height of the football is, [tex]h_{m}[/tex] = 7.39 m
The x velocity of the football is, Vₓ = 15.97 m/s
Hence, the distance traveled in the x direction is, S = 39.25 m
Explanation:
Given data,
The projected angle of the football to the horizontal, Ф = 37°
The velocity of the ball, v = 20 m/s
The initial velocity of the ball is, u = 20 m/s
The amount of time in air,
[tex]t_{f}=\frac{2uSin\phi}{g}[/tex]
[tex]t_{f}=\frac{2\times20Sin37}{9.8}[/tex]
[tex]t_{f}[/tex] = 2.45 s
Hence, the amount of the time in air, [tex]t_{f}[/tex] = 2.45 s
The maximum height,
[tex]h_{m}=\frac{u^{2}Sin^{2}\phi}{2g}[/tex]
[tex]h_{m}=\frac{20^{2}Sin^{2}37}{2\times9.8}[/tex]
[tex]h_{m}[/tex] = 7.39 m
Hence, the maximum height of the football is, [tex]h_{m}[/tex] = 7.39 m
The initial horizontal velocity,
Vₓ = V cos Ф
= 20 x 0.7986
= 15.97 m/s
The x velocity of the football is, Vₓ = 15.97 m/s
The distance traveled in the x direction
[tex]S=\frac{u^{2}Sin2\phi}{g}[/tex]
[tex]S=\frac{20^{2}Sin2(35)}{9.8}[/tex]
S = 39.25 m
Hence, the distance traveled in the x direction is, S = 39.25 m
