2: When applying the equations of kinematics for an ob-ject moving in one dimension, which of the following state-ments must be true?(a) The velocity of the object must remain constant.(b) The acceleration of the object must remain constant.(c) The velocity of the object must increase with time.6(d) The position of the object must increase with time.(e) The velocity of the object must always be in the samedirection as its acceleration.Q13: A cannon shell is fired straight up from the ground atan initial speed of 225 m/s. After how much time is the shellat a height of 6.20 × 102 m above the ground and movingdownward?(a) 2.96 s (b) 17.3 s (c) 25.4 s (d) 33.6 s (e) 43.0 sQ14: An arrow is shot straight up in the air at an initialspeed of 15.0 m/s. After how much time is the arrow movingdownward at a speed of 8.00 m/s?(a) 0.714 s (b) 1.240 s (c) 1.870 s (d) 2.350 s (e) 3.220sQ15: A rock is thrown downward from the top of a 40.0-m-tall tower with an initial speed of 12 m/s. Assuming negligibleair resistance, what is the speed of the rock just before hittingthe ground?(a) 28 m/s (b) 30 m/s (c) 56 m/s (d) 784 m/s (e) NoneQ16: On another planet, a marble is released from rest atthe top of a high cliff. It falls 4.00 m in the first 1 s of itsmotion. Through what additional distance does it fall in thenext 1 s?7

Sure, I can help with that. Here are the answers to your questions:

Q2: The correct answer is (b) The acceleration of the object must remain constant. This is because the equations of kinematics are derived under the assumption of constant acceleration.

Q13: To solve this, we can use the equation of motion: h = ut + 0.5gt^2, where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time. We know that u = 225 m/s, g = -9.8 m/s^2 (since it's acting downwards), and h = 620 m. We can solve this equation for t when the shell is at the peak of its trajectory (velocity = 0), and then when it's on its way down and at a height of 620 m. The correct answer is (b) 17.3 s.

Q14: We can use the equation v = u + gt, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time. We know that u = 15 m/s, v = -8 m/s (since it's moving downwards), and g = -9.8 m/s^2. Solving this equation for t gives us the correct answer of (c) 1.870 s.

Q15: We can use the equation v^2 = u^2 + 2gh, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and h is the height. We know that u = 12 m/s, g = 9.8 m/s^2, and h = 40 m. Solving this equation for v gives us the correct answer of (c) 56 m/s.

Q16: On this other planet, the marble falls 4 m in the first second. This implies that the acceleration due to gravity on this planet is 2*4 = 8 m/s^2 (since h = 0.5gt^2). In the next second, the marble will fall an additional distance of h = 0.5g(2)^2 = 16 m (since the time is now 2 seconds). So, the marble falls an additional 16 - 4 = 12 m in the next second.