A ball is thrown upward. After reaching a maximum height, it continues falling back to- ward Earth. On the way down, the ball is caught at the same height at which it was thrown upward.If the time (up and down) the ball remains in the air is 2.3 s, find its speed when it caught. The acceleration of gravity is 9.8 m/s2 . Ne- glect air resistance.

Answer in units of m/s

:

4.3 m/s

To solve this problem, we will use the kinematic equations of motion.

First, we need to find the time it takes for the ball to reach its maximum height when thrown upward. Since we are neglecting air resistance, we know that the initial vertical velocity when the ball is thrown upward is equal in magnitude but opposite in direction to the final vertical velocity when the ball is caught. This means that the time it takes for the ball to reach its maximum height is equal to the time it takes for the ball to fall from its maximum height to the catch height.

Let's call the time it takes for the ball to reach its maximum height "t". So, the time it takes for the ball to fall from the maximum height to the catch height is also "t".

We are given that the total time the ball remains in the air (up and down) is 2.3 seconds. Therefore, the equation becomes:

2t = 2.3

Solving for t:

t = 2.3 / 2
t = 1.15 seconds

Now, let's find the speed of the ball when it was thrown upward. We know that the initial vertical velocity is equal to the final vertical velocity when the ball is caught, but in opposite directions. The final vertical velocity is 0 m/s (at the highest point), so the initial vertical velocity when the ball is thrown upward is also 0 m/s.

Using the equation for vertical velocity:

v = u + at

where:
v is the final velocity (0 m/s)
u is the initial velocity (0 m/s)
a is the acceleration due to gravity (-9.8 m/s^2) in the downward direction
t is the time (1.15 seconds)

0 = 0 - (9.8)(1.15)

Solving for u:

u = (9.8)(1.15)
u = 11.27 m/s

So, the initial vertical velocity (speed) when the ball is thrown upward is 11.27 m/s.

Finally, we need to find the speed of the ball when it was caught. Since the initial vertical velocity when the ball is thrown upward is equal in magnitude but opposite in direction to the final vertical velocity when the ball is caught, we just need to take the absolute value of the initial vertical velocity. Therefore:

Speed when caught = |11.27| = 11.27 m/s

Hence, the speed of the ball when it was caught is 11.27 m/s.

To find the speed of the ball when it is caught, we can first calculate the time it takes for the ball to reach its maximum height.

We know that the time it takes for the ball to reach its maximum height is equal to half of the total time it remains in the air. So, the time it takes for the ball to reach its maximum height is 2.3 seconds divided by 2, which is 1.15 seconds.

The next step is finding the velocity of the ball at its maximum height. We can use the equation:

V = U + at

Where:
V = final velocity (unknown)
U = initial velocity (which is the velocity when the ball was caught)
a = acceleration due to gravity (-9.8 m/s^2)
t = time (1.15 seconds)

Since the ball is thrown upwards, the initial velocity is positive. Therefore, the equation becomes:

V = U + (-9.8 m/s^2)(1.15 s)

Simplifying this equation, we can say:

V = U - 11.27 m/s

Now, we need to consider the time it takes for the ball to fall back to the same height at which it was thrown upwards, which is also 1.15 seconds.

Using the equation of motion, we can calculate the final velocity of the ball when it is caught:

V = U + at

Where:
V = final velocity (unknown)
U = initial velocity (V in the previous equation)
a = acceleration due to gravity (9.8 m/s^2)
t = time (1.15 seconds)

So, the equation becomes:

V = U + (9.8 m/s^2)(1.15 s)

Substituting the value of U from the previous equation, we have:

V = (U - 11.27 m/s) + (9.8 m/s^2)(1.15 s)

Simplifying this equation, we get:

V = U + 11.26 + 11.27
V = U + 22.53 m/s

Since we know that the initial velocity U is equal to the speed when the ball is caught, we can rewrite the equation as:

V = V + 22.53 m/s

Simplifying further, we find that:

2V = 22.53 m/s

Dividing both sides of the equation by 2, we get:

V = 11.26 m/s

Therefore, the speed of the ball when it is caught is 11.26 m/s.