Suppose you throw a ball vertically upward with a speed of 49 m/s. Neglecting air friction, what would be the speed of the ball 2.0 second later?
29.4??
you got the correct result ... so I'm guessing you used the right technique
To find the speed of the ball 2.0 seconds later, we need to take into account the acceleration due to gravity.
The acceleration due to gravity is approximately 9.8 m/s², and since the ball is thrown vertically upward, it will decelerate at this rate until it reaches its highest point and starts falling back down.
To find the speed of the ball after 2.0 seconds, we can use the following equation of motion:
v = u + at
Where:
v is the final velocity
u is the initial velocity
a is the acceleration due to gravity
t is the time
Given:
u = 49 m/s
a = -9.8 m/s² (negative because it acts in the opposite direction to the initial velocity)
t = 2.0 s
Plugging in the values into the equation, we have:
v = 49 m/s + (-9.8 m/s²) × 2.0 s
Simplifying:
v = 49 m/s - 19.6 m/s
v = 29.4 m/s
Therefore, the speed of the ball 2.0 seconds later would be 29.4 m/s.
To find the speed of the ball 2.0 seconds later, we need to use the equations of motion. Specifically, we can use the equation for vertical free-fall motion, which states that the final velocity of an object in free-fall is given by:
v = u + at
Where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration due to gravity (which we assume to be -9.8 m/s^2, negative because it acts in the opposite direction)
- t is the time
Given:
- u = 49 m/s (upward)
- t = 2.0 s
Substituting the given values into the equation, we get:
v = 49 m/s + (-9.8 m/s^2) * 2.0 s
Simplifying the calculation, we have:
v = 49 m/s - 19.6 m/s
Finally, we find:
v = 29.4 m/s
Therefore, the speed of the ball 2.0 seconds later would be 29.4 m/s.