A soccer ball is kicked with an initial horizontal velocity of 19.0 m/s and an initial vertical velocity of 14.0 m/s
6) How high above the ground is the ball 1.1 seconds after it is kicked?
h(t) = 14.0t - 4.9t^2
h(1.1) = 9.47m
To find the height of the ball above the ground at a certain time, you need to use the equations of motion. In this case, you can use the kinematic equation for vertical motion:
h = h₀ + v₀t + (1/2)gt²
Where:
- h is the height of the ball above the ground at time t
- h₀ is the initial height (which is 0 since the ball starts from the ground)
- v₀ is the initial vertical velocity (14.0 m/s)
- t is the time (1.1 seconds)
- g is the acceleration due to gravity (-9.8 m/s²)
Now, let's plug in the values and calculate the height of the ball:
h = 0 + (14.0 m/s)(1.1 s) + (1/2)(-9.8 m/s²)(1.1 s)²
First, multiply 14.0 m/s by 1.1 s to get 15.4 m/s.
Next, square 1.1 s to get 1.21 s².
Finally, multiply (1/2)(-9.8 m/s²) by 1.21 s² and get -5.59 m.
Now, let's add all the values together:
h = 0 + 15.4 m + (-5.59 m)
h = 9.81 m
So, the ball is approximately 9.81 meters above the ground 1.1 seconds after it was kicked.