A football is kicked from ground level with an initial velocity of 20.4 m/s at angle of 47.0° above the horizontal. How long, in seconds, is the football in the air before it hits the ground? Ignore air resistance.
just find when the height is zero
20.4 sin47° t - 4.9t^2 = 0
t = (20.4 sin47°)/4.9
To find the time the football is in the air before it hits the ground, we can use the equation of motion in the vertical direction:
h = (viy)t + (1/2)gt^2
Where:
h = height (which is zero since it starts and ends at ground level)
viy = initial vertical velocity
t = time
g = acceleration due to gravity (approximately 9.8 m/s^2)
In this case, the initial vertical velocity (viy) can be found by multiplying the initial velocity (20.4 m/s) by the sine of the angle (47.0°).
viy = (20.4 m/s) * sin(47.0°)
Now, we can rewrite the equation for h as:
0 = (20.4 m/s * sin(47.0°)) * t + (1/2) * 9.8 m/s^2 * t^2
This is a quadratic equation. We can solve for t by rearranging the equation:
0 = (20.4 * sin(47.0)) * t + (1/2) * (9.8) * t^2
0 = (20.4 * sin(47.0)) * t + 4.9 * t^2
Now we can solve this quadratic equation. Let's imagine it as:
at^2 + bt + c = 0
where:
a = 4.9
b = 20.4 * sin(47.0)
c = 0
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values:
t = (-(20.4 * sin(47.0)) ± √((20.4 * sin(47.0))^2 - 4 * 4.9 * 0)) / (2 * 4.9)
Evaluating this equation gives us two possible times. However, since we are only interested in the time it takes for the football to hit the ground, we can ignore any negative solutions or values of t that indicate the time it takes for the ball to reach its peak height. Therefore, we only consider the positive value of t.
Calculating this with a calculator or software, we find that the positive solution is approximately 2.25 seconds. Therefore, the football is in the air for approximately 2.25 seconds before it hits the ground.