A player kicks a soccer ball from ground level and sends it flying at an angle of 30 degrees at a speed of 26 m/s. How long was the ball in the air? Round the answer to the nearest tenth of a second.
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A soccer player kicking a ball; the ball soaring through the air and landing on the ground
To find out how long the ball was in the air, we can use the kinematic equations of motion. Specifically, we can use the equation that relates the vertical displacement (Δy), initial vertical velocity (Vy₀), time (t), and acceleration due to gravity (g).
The equation is:
Δy = Vy₀ * t - (1/2) * g * t²
Since the ball was kicked from ground level, the initial vertical displacement (Δy) is zero. We can set this to zero and solve for time (t):
0 = Vy₀ * t - (1/2) * g * t²
In this case, the initial vertical velocity (Vy₀) can be calculated by multiplying the initial speed (26 m/s) by the sine of the launch angle (30 degrees):
Vy₀ = 26 m/s * sin(30°)
To use this equation, we also need to know the value of the acceleration due to gravity (g), which is approximately 9.8 m/s².
Now, let's substitute these values into the equation and solve for time (t):
0 = (26 m/s * sin(30°)) * t - (1/2) * (9.8 m/s²) * t²
Simplifying the equation:
0 = (13 m/s) * t - (4.9 m/s²) * t²
This is a quadratic equation in terms of time (t). We can solve it by factoring or using the quadratic formula. However, in this case, we'll use factoring:
0 = t * ((13 m/s) - (4.9 m/s²) * t)
Setting each factor equal to zero:
t = 0
or
(13 m/s) - (4.9 m/s²) * t = 0
Solving the second equation for time (t):
(4.9 m/s²) * t = 13 m/s
t = 13 m/s / (4.9 m/s²)
t ≈ 2.653 seconds
Therefore, the ball was in the air for approximately 2.7 seconds (rounded to the nearest tenth of a second).
A player kicks a soccer ball from ground level and sends it flying at an angle of 30 degrees at a speed of 26 m/s. How long was the ball in the air? Round the answer to the nearest tenth of a second.
hf=hi+vi'*t-4.9t^2
hf, hi =0
Vi'=25*sin30deg. solve for t.