When a ball is thrown upward, it experiences a downward accelera-

tion of magnitude 9.8 m/s
2
, neglecting air resistance. With what
velocity must a ball leave a thrower’s hand in order to climb for 2.2 s
before stopping?

V = Vo + g*t

V = 0
Vo = Initial velocity.
g = -9.8 m/s^2
t = 2.2 s.
Solve for Vo.

To find the velocity with which the ball must leave the thrower's hand, we can use the kinematic equation:

Vf = Vi + at

Where:
- Vf is the final velocity
- Vi is the initial velocity
- a is the acceleration
- t is the time

In this case, the final velocity (Vf) will be 0 m/s since the ball stops at its highest point.

The acceleration (a) is given as -9.8 m/s^2 because it is in the opposite direction of the motion. The negative sign indicates that the acceleration is acting downwards.

The time (t) is given as 2.2 seconds, which is the duration for the ball to climb.

Now, we can substitute the given values into the equation and solve for the initial velocity (Vi):

0 = Vi - 9.8 * 2.2

0 = Vi - 21.56

Vi = 21.56 m/s

Therefore, the ball must be thrown with an initial velocity of 21.56 m/s in order to climb for 2.2 seconds before stopping.