Consider the position of a ball thrown down
with an initial speed of 17 m/s.What will be its position after 2.6 s? Let
the initial position be 0. The acceleration of
gravity is 9.8 m/s
2
.
Answer in units of m
216
To determine the position of the ball after 2.6 seconds, you can use the kinematic equation:
\(d = d_0 + v_0t + \frac{1}{2}gt^2\)
Where:
- \(d\) is the final position
- \(d_0\) is the initial position (which is 0 in this case)
- \(v_0\) is the initial velocity (17 m/s in this case)
- \(t\) is the time (2.6 s in this case)
- \(g\) is the acceleration due to gravity (9.8 m/s^2 in this case)
Plugging the given values into the equation, we get:
\(d = 0 + (17 \, \text{m/s})(2.6 \, \text{s}) + \frac{1}{2}(9.8 \, \text{m/s}^2)(2.6 \, \text{s})^2\)
Simplifying, we have:
\(d = 0 + 44.2 \, \text{m} + \frac{1}{2}(9.8 \, \text{m/s}^2)(6.76 \, \text{s}^2)\)
\(d = 0 + 44.2 \, \text{m} + 32.728 \, \text{m}\)
\(d = 76.928 \, \text{m}\)
Therefore, the position of the ball after 2.6 seconds will be 76.928 meters.