a ball is launched upward from the ground at 30.0 m/s. its location at 1 - second intervals is shown…
a ball is launched upward from the ground at 30.0 m/s. its location at 1 - second intervals is shown. determine the time, velocity, and acceleration at the four indicated locations. enter negative values for any downward - directed vector.\n\n| time (s) | velocity (m/s) | accelen (m/s/s) |\n| ---- | ---- | ---- |\n| a | 0.0 | 30.0 | -9.81 |\n| c | 2 | | -9.81 |\n| d | 3 | 0 | -9.81 |\n| f | 5 | tap me! | -9.81 |\n\ncolor - coding key:\nnot checked correct incorrect entering answer\ncheck answers
Answer
Explanation:
Step1: Use velocity - time formula
The formula for velocity $v = v_0+at$, where $v_0 = 30.0$ m/s is the initial velocity, $a=- 9.81$ m/s² is the acceleration due to gravity.
Step2: Calculate velocity at $t = 2$s
Substitute $t = 2$s, $v_0 = 30.0$ m/s and $a=-9.81$ m/s² into the formula $v=v_0 + at$. So $v=30.0+( - 9.81)\times2=30.0 - 19.62 = 10.38$ m/s.
Step3: Calculate velocity at $t = 5$s
Substitute $t = 5$s, $v_0 = 30.0$ m/s and $a=-9.81$ m/s² into the formula $v=v_0+at$. So $v=30.0+( - 9.81)\times5=30.0-49.05=-19.05$ m/s.
Answer:
| Time (s) | Velocity (m/s) | Acceleration (m/s²) |
|---|---|---|
| A: 0.0 | 30.0 | - 9.81 |
| C: 2 | 10.38 | - 9.81 |
| D: 3 | 0 | - 9.81 |
| F: 5 | - 19.05 | - 9.81 |