A ball is thrown straight up from a bridge at a speed of 11.0 m/s. What will be its velocity (speed and direction) after 2.0 seconds?
v(t) = v0 + at
remember that the acceleration is downward (negative)
that is correct
is a gravity? 9.80m/s^2
thanks
To find the velocity of the ball after 2.0 seconds, we need to consider the motion of the ball as it travels upwards from the bridge.
When the ball is thrown straight up, it experiences a negative acceleration due to gravity. This means that its velocity decreases over time.
To calculate the velocity of the ball after 2.0 seconds, we can use the following equation:
v = u + at
Where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time
Given:
- u = 11.0 m/s (initial velocity)
- t = 2.0 seconds (time)
We know that the acceleration due to gravity is approximately -9.8 m/s² (negative because it opposes the motion).
Substituting the values into the equation, we have:
v = 11.0 m/s + (-9.8 m/s²) * 2.0 s
Simplifying the equation, we get:
v = 11.0 m/s - 19.6 m/s
v = -8.6 m/s
Therefore, the velocity of the ball after 2.0 seconds will be -8.6 m/s, indicating that it is moving downwards.