Billy kicks a soccer ball with an initial velocity of 24 ft/sec into the goal. How long was the ball in the air?
If the goalie caught the ball at a height of 4 feet from the ground, how long was the ball in the air before it was caught?
To find out how long the ball was in the air, you need to use the kinematic equation for vertical motion. The equation is:
h = v₀t + (1/2)gt²
Where:
- h is the height of the ball at a specific time.
- v₀ is the initial velocity of the ball.
- t is the time.
- g is the acceleration due to gravity (which is approximately 32 ft/sec²).
For the first question:
1. The initial velocity of the ball, v₀, is given as 24 ft/sec.
2. The height of the ball, h, is not mentioned. However, we assume that the ball was kicked from the ground level, so its initial height is 0 ft.
3. We're looking for the time, t.
Plugging in the values into the equation, we get:
0 = (24)t + (1/2)(32)t²
Simplifying the equation:
0 = 24t + 16t²
Now, we can solve the quadratic equation to find the values of t.
For the second question:
1. The initial velocity, v₀, remains the same as 24 ft/sec.
2. The height of the ball when it was caught, h, is given as 4 ft.
3. We'll calculate the time, t, again.
Plug in the values into the equation:
4 = 24t + (1/2)(32)t²
Simplifying the equation:
0 = 32t² + 24t - 4
Solving this quadratic equation will give us the time, t.
I will calculate the values of t for both scenarios and provide the answers.