To solve this problem, we will use the kinematic equation for uniformly accelerated motion, as gravity is the only force acting on the ball (neglecting friction). The equation we will use is:

$v = v_0 + at$

where:

• $v$ is the final velocity,
• $v_0 = 0.50 \, \text{m/s}$ is the initial velocity (speed at the moment the ball is released),
• $a = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity,
• $t = 0.70 \, \text{seconds}$ is the time after the ball is released.

Now, plug in the values:

$v = 0.50 \, \text{m/s} + (9.8 \, \text{m/s}^2)(0.70 \, \text{s})$

$v = 0.50 \, \text{m/s} + 6.86 \, \text{m/s}$

$v = 7.36 \, \text{m/s}$

Rounding to two significant figures (as in the given data), we get:

$v \approx 7.4 \, \text{m/s}$

Thus, the speed of the ball 0.70 seconds after it is released is 7.4 m/s.

Answer: 7.4 m/s

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Would you like more details or have any further questions on this?

Here are some additional related questions:

1. What would be the velocity if the ball were thrown upwards instead?
2. How long would it take for the ball to hit the ground if thrown from the same height?
3. What is the displacement of the ball after 0.70 seconds?
4. How would air resistance affect the result if it were not neglected?
5. How does the mass of the ball affect its motion under gravity?

Tip: Remember, when dealing with gravity near Earth's surface, the acceleration due to gravity is always approximately $9.8 \, \text{m/s}^2$.