A ball is thrown vertically into the air and when it returns after an interval of 2 seconds it is caught. What is the acceleration of the ball after leaving the hand?
I have no idea how to solve this!
You have not been given enough information to answer this problem.
In general, for objects falling near the surface of the earth, the acceleration is -g = -9.8 m/s^2, no matter how long it takes to come down. The minus sign indicates downward acceleration
If you did not know the value of g (for example, if you were on an unknown planet), you could determine the acceleration (a) if you knew the launch velocity of the ball (V). The time it takes to come down (T) is given by
a T/2 = V
To solve this problem, we can use the equations of motion for objects in free fall.
First, we need to understand that when the ball is at its highest point, its velocity will be zero. We can use this information to figure out the time it takes to reach the highest point.
Since the ball takes 2 seconds to return to the hand, we know that it takes 1 second to reach the highest point because the time taken will be the same for both the upward and downward journey.
Now, we can use the equation for displacement in the vertical direction during free fall:
h = (1/2) * g * t^2
where h is the displacement (in this case, the height reached), g is the acceleration due to gravity, and t is time.
We know that the ball returns to the same height from which it was thrown, so the displacement is zero.
0 = (1/2) * g * (1^2)
Simplifying the equation, we find:
0 = (1/2) * g
Therefore, g = 0
The acceleration of the ball after leaving the hand is zero. This means that the ball is not experiencing any additional acceleration other than that due to gravity, which is 9.8 m/s^2.