a person throws a baseball into the air with an initial vertical velocity of 30 feet per second and then lets the ball hit the ground. The ball is released 5 feet above the ground. How long is the ball in the air?
h=-16t^2 + Vot + h (equation)
5=-16t^2 + 30t + 5
subtract 5
0 = -16t^2 + 30t
factor out a t
0=t (-16t + 30)
0=t 0r -16t + 30=0
13 / 16
approximately 1.87 second
To find out how long the ball is in the air, we can use the kinematic equation for vertical motion:
h = h0 + v0t - (1/2)gt^2
Where:
- h is the final height (0 in this case, since the ball hits the ground)
- h0 is the initial height (5 feet)
- v0 is the initial velocity (30 feet/second)
- g is the acceleration due to gravity (-32.2 feet/second^2, assuming Earth's gravity)
Since the ball hits the ground, we can set h equal to 0:
0 = 5 + 30t - (1/2)(-32.2)t^2
Simplifying:
0 = 5 + 30t + 16.1t^2
Rearranging the equation:
16.1t^2 + 30t + 5 = 0
Now we can use the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 16.1, b = 30, and c = 5. Plugging in these values:
t = (-30 ± √(30^2 - 4 * 16.1 * 5)) / (2 * 16.1)
Simplifying:
t = (-30 ± √(900 - 322)) / 32.2
t = (-30 ± √578) / 32.2
t ≈ (-30 ± 24.04) / 32.2
Now, we have two possible solutions for t:
1. t = (-30 + 24.04) / 32.2 ≈ -0.19 seconds (which we discard since time cannot be negative in this context).
2. t = (-30 - 24.04) / 32.2 ≈ -1.03 seconds (which we also discard since time cannot be negative in this context).
Therefore, the ball is in the air for approximately 1.03 seconds.
To find out how long the ball is in the air, we can use the equations of motion for an object in freefall. The equation we'll be using is:
h = ut + (1/2)gt^2
Where:
- h is the height of the object
- u is the initial velocity in the vertical direction
- g is the acceleration due to gravity (which is approximately 32 feet per second squared)
- t is the time
In this case, the ball is thrown upwards and then falls back to the ground, so we need to consider both the upward and downward motion.
First, let's find the time it takes for the ball to reach its highest point. At the highest point, the vertical velocity would be zero. We can use the equation of motion:
u = v + gt
Since the initial vertical velocity (u) is 30 feet per second and the final vertical velocity (v) is zero, we can rearrange the equation and solve for t:
t = (v - u) / g
t = (0 - 30) / (-32)
t = 30 / 32
t ≈ 0.9375 seconds
Now, to find the total time the ball is in the air, we need to consider both the upward motion and the downward motion. Since the ball took 0.9375 seconds to reach its highest point, it will take the same amount of time to fall back down to the release point.
So, the total time the ball is in the air is:
Total time = Time to reach highest point + Time to fall back down
Total time = 2 * Time to reach highest point
Total time = 2 * 0.9375
Total time ≈ 1.875 seconds
Therefore, the ball is in the air for approximately 1.875 seconds.
hf=ho+vi*t-g/2 * t^2
0=5+30t-4.9t^2
solve the quadratic with the quadratic equation.