practice/lab
directions: answer the following questions based on our notes.
part i - conceptual
a student throws a ball straight upward and it takes 3 seconds to reach the top. answer the following questions.

question | answer
1. what is the velocity of the ball at the very top? explain why! |
2. what is the acceleration of the ball going up? explain why! |
3. is the acceleration of the ball going down the same or different as the acceleration going up? explain why! |
4. is the ball faster when the student throws the ball or right before the ball hits his hand coming back down? explain why! |
does it take longer for the ball to be thrown upward or come back down? explain why! |

Question

practice/lab
directions: answer the following questions based on our notes.
part i - conceptual
a student throws a ball straight upward and it takes 3 seconds to reach the top. answer the following questions.

question | answer
1. what is the velocity of the ball at the very top? explain why! |
2. what is the acceleration of the ball going up? explain why! |
3. is the acceleration of the ball going down the same or different as the acceleration going up? explain why! |
4. is the ball faster when the student throws the ball or right before the ball hits his hand coming back down? explain why! |
does it take longer for the ball to be thrown upward or come back down? explain why! |

Accepted Answer

### Question 1
# Brief Explanations:
At the very top of its trajectory, the ball momentarily stops changing its vertical position. Velocity is the rate of change of position. Since the ball is not moving up or down at that instant, its velocity is 0 m/s.
# Answer:
The velocity of the ball at the very top is 0 m/s. This is because at the peak of its motion, the ball momentarily comes to rest before starting to fall back down, so the rate of change of its position (velocity) is zero.

### Question 2
# Brief Explanations:
When the ball is going up, the only significant force acting on it (neglecting air resistance) is gravity. Gravity causes an acceleration towards the center of the Earth, which is approximately $ - 9.8\space m/s^{2}$ (negative if we take upward as positive) or $9.8\space m/s^{2}$ downward. So the acceleration of the ball going up is equal to the acceleration due to gravity, $g = 9.8\space m/s^{2}$ (directed downward).
# Answer:
The acceleration of the ball going up is $9.8\space m/s^{2}$ (or $ - 9.8\space m/s^{2}$ if upward is positive) directed downward. This is because the force of gravity is acting on the ball, and according to Newton's second law, $F = ma$, the acceleration due to gravity ($g$) is the acceleration of the ball (neglecting air resistance) as gravity is the net force acting on it during upward motion.

### Question 3
# Brief Explanations:
When the ball is going down, the only significant force acting on it (neglecting air resistance) is still gravity. The acceleration due to gravity is a constant near the Earth's surface and acts downward. So whether the ball is moving up or down (neglecting air resistance), the acceleration is the same in magnitude ($9.8\space m/s^{2}$) and direction (downward). If air resistance is considered, the acceleration going down would be slightly less, but in most basic physics problems, air resistance is neglected, so the acceleration is the same.
# Answer:
The acceleration of the ball going down is the same as the acceleration going up (neglecting air resistance). This is because the net force acting on the ball (gravity, neglecting air resistance) is the same in both cases, and according to Newton's second law ($F = ma$), the acceleration (which is $g$, the acceleration due to gravity) remains constant in magnitude and direction (downward) for the motion of the ball near the Earth's surface.

### Question 4
# Brief Explanations:
Assuming no air resistance, the speed of the ball when the student throws it (initial speed) and the speed right before it hits his hand coming back down is the same. This is due to the conservation of mechanical energy. The ball starts with a certain kinetic energy (and potential energy at the hand level) and as it goes up, kinetic energy is converted to potential energy. When it comes back down, the potential energy is converted back to kinetic energy (neglecting air resistance, so no energy is lost). So the speed when thrown and the speed when returning to the hand (at the same height) should be equal. If air resistance is considered, the ball would lose some energy due to air resistance, so it would be slower when coming back down. But in a basic physics context without air resistance, they are equal.
# Answer:
(Assuming no air resistance) The ball has the same speed when the student throws it and right before it hits his hand coming back down. This is because mechanical energy (kinetic + potential) is conserved (neglecting air resistance). When thrown, the ball has kinetic energy (and potential energy at the hand's height). As it rises, kinetic energy converts to potential energy, and as it falls back, potential energy converts back to kinetic energy. At the same height (the hand's height), the potential energy is the same as initially, so the kinetic energy (and thus speed, since $KE=\frac{1}{2}mv^{2}$) is the same. (If air resistance is considered, the ball is faster when thrown, as air resistance dissipates energy, reducing the speed on the way down.)

### Question 5
# Brief Explanations:
(Assuming no air resistance) The time to go up and the time to come back down are the same. When going up, the initial velocity $u$ decreases to 0 at the top with an acceleration $ - g$ (taking upward as positive). Using the equation $v=u + at$, at the top $v = 0$, so $0=u - gt_{up}$, so $t_{up}=\frac{u}{g}$. The maximum height $h$ reached is given by $v^{2}-u^{2}=2ah$, so $0 - u^{2}=- 2gh$, so $h=\frac{u^{2}}{2g}$. When coming down, the ball starts from rest ($u = 0$) at height $h$ and accelerates downward with $a = g$. Using the equation $h=ut+\frac{1}{2}at^{2}$, $h = 0+\frac{1}{2}gt_{down}^{2}$, so $t_{down}=\sqrt{\frac{2h}{g}}$. Substituting $h=\frac{u^{2}}{2g}$ into the equation for $t_{down}$, we get $t_{down}=\sqrt{\frac{2\times\frac{u^{2}}{2g}}{g}}=\sqrt{\frac{u^{2}}{g^{2}}}=\frac{u}{g}$, which is equal to $t_{up}$. If air resistance is considered, the time to come down is longer because air resistance opposes the motion when going up (reducing the maximum height and the time to go up) and also opposes the motion when coming down but the average speed on the way down is less than the average speed on the way up (since the initial speed up is higher than the final speed down), leading to a longer time to come down. But in a basic physics context without air resistance, the times are equal.
# Answer:
(Assuming no air resistance) It takes the same amount of time for the ball to be thrown upward and to come back down. This is because the motion is symmetric. The time to reach the top (where velocity becomes zero) is determined by the initial velocity and acceleration due to gravity, and the time to fall back from that height (with initial velocity zero and the same acceleration due to gravity) is equal to the time to go up, as shown by the equations of motion. (If air resistance is considered, it takes longer to come down because air resistance reduces the speed on the way down more significantly compared to its effect on the way up.)

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