To determine the velocity of the ball just before it strikes the boy's hand, we can use the equation of motion for vertical motion under gravity. Since the ball is thrown vertically upward, we assume that the acceleration due to gravity acts in the opposite direction.
Given: Initial velocity, u = 20 m/s (upward) Acceleration due to gravity, g = 10 m/s² (downward)
The final velocity just before the ball strikes the boy's hand is 0 m/s since it momentarily comes to rest at its highest point before starting to fall back down. Therefore, we need to find the time it takes for the ball to reach its highest point.
Using the equation of motion for vertical velocity: v = u + gt
At the highest point, the final velocity v is 0 m/s, so we have: 0 = 20 + (-10)t
Simplifying the equation: -10t = -20
Dividing both sides by -10: t = -20 / -10 t = 2 seconds
It takes 2 seconds for the ball to reach its highest point. Now, we can find the velocity of the ball just before it strikes the boy's hand. We use the equation:
v = u + gt
Substituting the values: v = 20 + (-10)(2) v = 20 - 20 v = 0 m/s
Therefore, the velocity of the ball just before it strikes the boy's hand is 0 m/s, indicating that it momentarily comes to rest at the highest point of its trajectory.