Respuesta :
Answer:
The correct option is;
The kick is good! The ball clears the crossbar by nearly 15 feet.
Step-by-step explanation:
Here we have;
Velocity of ball = 70 ft/s
Angle of motion = 50°
Therefore;
Vertical component of velocity, [tex]v_y[/tex] = v×sinθ = v×sin×50 = 53.62 ft/s
[tex]v_y[/tex] = u[tex]_y[/tex] - g·t
At maximum height [tex]v_y[/tex] = 0, therefore;
u[tex]_y[/tex] = g·t and
t = u[tex]_y[/tex]/g
Where:
u[tex]_y[/tex] = Initial vertical velocity = 53.62 ft/s
g = Acceleration due to gravity = 32.1740 ft/s²
∴ t = 53.62/32.1740 = 1.66666 s
Total time of flight = 2 × 1.66666 = 3.33332 s
Max height is given by the following relation;
v[tex]_y[/tex]² = u[tex]_y[/tex]² - 2·g·s
v[tex]_y[/tex] = 0 ft/s at maximum height, therefore;
u[tex]_y[/tex]² = 2·g·s
53.62² = 2×32.1740×s
s = 53.62² ÷(2×32.1740) = 44.69 ft
Horizontal velocity = 70×cos(50) = 44.995 ft/s
Hence the time it takes the ball to reach 42 yards is given by the following relation;
42 yards = 126 ft
[tex]Time, t = \frac{Distance}{Velocity} = \frac{126}{44.995} = 2.8 \, s[/tex]
Height of ball at 2.8 s is given by;
Direction of ball at 2.8 s = downwards
Hence time after peak = 2.8 - 1.666 = 1.1336429 s
Position at 2.8 s is given by
s = u[tex]_y[/tex] ·t + 1/2·g·t²
u[tex]_y[/tex] = 0 ft/s as ball is on second half of flight
∴ s = 1/2·g·t² = 1/2×32.1740×1.1336429²= 20.674 ft below maximum height
∴ Height above ground = Maximum height - 20.674 ft = 44.69 - 20.674
Height above ground = 24.01 ft
Hence the ball clears the crossbar by 24.01 - 10 = 14.01 ft
The best option is therefore;
The kick is good! The ball clears the crossbar by nearly 15 feet.