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
To find the position of the ball after 2.6 seconds, we can use the kinematic equation:
s = ut + (1/2)at^2
where:
- s is the position (displacement)
- u is the initial velocity
- t is the time
- a is the acceleration
In this case, the ball is thrown downward, so the initial velocity (u) is 17 m/s and the acceleration (a) is the acceleration due to gravity, which is -9.8 m/s^2 (negative because the ball is moving in the downward direction).
Plugging these values into the equation, we get:
s = (17 m/s)(2.6 s) + (1/2)(-9.8 m/s^2)(2.6 s)^2
Calculating this expression:
s = 44.2 m + (1/2)(-9.8 m/s^2)(6.76 s^2)
= 44.2 m + (-3.788 m)
= 40.412 m
Therefore, the position of the ball after 2.6 seconds will be approximately 40.412 meters.