a. When a ball is thrown straight up, by how much does the speed decrease each second?
b. After the ball reaches the top and begins its return back down, by how much does its speed increase each second?
c. Compare the times going up and coming down.
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a. To determine how much the speed of the ball decreases each second when thrown straight up, we need to consider the effect of gravity.
The acceleration due to gravity is constant and acts in the opposite direction of the ball's motion when it is thrown up. The commonly accepted acceleration due to gravity is approximately 9.8 meters per second squared (m/s^2).
So, the speed of the ball decreases by 9.8 m/s every second it is in the air when thrown straight up.
b. Once the ball reaches the top of its trajectory and starts its descent, the speed of the ball actually increases each second due to the acceleration of gravity. Again, the acceleration due to gravity is approximately 9.8 m/s^2, but this time it acts in the same direction as the ball's motion when it is coming down.
Thus, the speed of the ball increases by 9.8 m/s every second it is in the air when coming back down.
c. The time it takes for the ball to reach its highest point and come back down to its initial launch height is the same. This is because the upward and downward motion of the ball are symmetrical.
To calculate the total time for the ball to reach the top and come back down, we can divide the time into two parts: the time going up and the time coming down.
The time going up would be determined by dividing the initial velocity of the ball (when it was thrown up) by the acceleration due to gravity.
The time coming down would be determined by dividing the final velocity of the ball (when it starts its descent) by the acceleration due to gravity.
Since the velocity when the ball starts descending is the same as the initial velocity when it was thrown up (just in the opposite direction), the time going up and the time coming down should be equal.