Ball is thrown vertically upward with a speed of 25.0 m/s. What is it's velocity when it returns to the level from which it had started ?
After computations: The ball reaches its highest pt at 32.4 meters.
It takes the ball 2.55sec to reach its highest pt.
It take the ball 2.55 sec to hit ground after it reaches its highest point
so what is its velocity ?
Use
v=u+at
where a=-9.8,
and
u=0 (at the highest point)
You could have arrived at the same conclusion by considerations of kinetic energy.
To find the velocity of the ball when it returns to the level from which it started, we can use the concept of projectile motion.
When the ball is thrown vertically upward, its initial velocity is 25.0 m/s. As the ball reaches its highest point, its velocity becomes zero. This is because at the highest point, the ball momentarily comes to a stop before it starts falling back down due to gravity.
Since the ball takes 2.55 seconds to reach its highest point, it will also take the same amount of time to reach the ground from the highest point. This means the total time for the ball to go up and come back down is 2.55 seconds.
Now, to calculate the velocity of the ball when it returns to the starting level, we can use the equation of motion:
final velocity = initial velocity + (acceleration x time)
Since the ball is moving down when it returns to the starting level, the acceleration due to gravity is in the downward direction and its value is approximately 9.8 m/s^2. The time taken to reach the ground from the highest point is 2.55 seconds.
So, plugging these values into the equation:
final velocity = 0 + (-9.8 m/s^2 x 2.55 s) = -25.0 m/s
The negative sign indicates that the velocity is in the downward direction, which means the ball is moving downwards when it returns to the starting level. So, the velocity of the ball when it returns to the level from which it started is -25.0 m/s.