A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 13m/s when the hand is 2.3m above the ground, how long does it stay in the air?
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To find out how long the ball stays in the air, we need to determine the time it takes for the ball to reach its highest point and then come back down to the student's hand.
To do this, we will use the equation of motion for vertical motion:
v = u + at
where:
- v is the final velocity (in this case, 0 m/s as the ball comes to rest at its highest point and when it returns to the student's hand),
- u is the initial velocity (13 m/s),
- a is the acceleration (in this case, the acceleration due to gravity, which is approximately 9.8 m/s^2),
- t is the time.
To find the time it takes for the ball to reach its highest point, we can rearrange the equation:
0 = 13 - 9.8t
Simplifying the equation, we get:
-9.8t = -13
Dividing both sides of the equation by -9.8:
t = 13 / 9.8
Calculating this value, we find that it takes approximately 1.33 seconds for the ball to reach its highest point.
To find the total time the ball stays in the air, we need to double this value, as the ball takes the same amount of time to come back down to the student's hand:
Total time = 2 * 1.33
Calculating this value, we find that the ball stays in the air for approximately 2.66 seconds.