A man leans over the edge of a cliff and throws a rock upward at 4.9 m/s. Neglecting air resistance, two seconds later the rock's speed is what?

Well, well, well, looks like we have a rockstar here! So this man throws a rock upward with a speed of 4.9 m/s. After two seconds, you want to know its speed, huh? Alright, let me calculate the rocket science for you... Just kidding, it's not rocket science, it's rock science!

Now, let's break it down. When the rock is thrown upward, it goes against gravity, which slows it down. In those two seconds, gravity has been giving it the ol' gravitational pull. So, let's calculate how much the speed has decreased.

Using a little physics magic, we know that the acceleration due to gravity is 9.8 m/s² (trust me, I didn't pull that number out of thin air, it's a gravitational constant). Multiplying this acceleration by the time of fall, which is 2 seconds, the rock's speed would have decreased by approximately 19.6 m/s.

But wait! The rock was thrown upwards, so the gravitational acceleration would actually decrease the speed by 19.6 m/s. So, if we subtract this from the initial speed of 4.9 m/s, we get a final speed of... drumroll, please... -14.7 m/s!

Yes, you heard it right, negative. The rock is moving downwards at 14.7 m/s after two seconds. Which means it's on a one-way ticket to go rock bottom!

To determine the rock's speed two seconds later, you can use the kinematic equation for projectile motion:

v_final = v_initial + (a * t)

Where:
- v_final is the final velocity of the rock
- v_initial is the initial velocity of the rock (4.9 m/s upward in this case)
- a is the acceleration due to gravity (-9.8 m/s^2 downward)
- t is the time elapsed (2 seconds)

Substituting the values into the equation:

v_final = 4.9 m/s - (9.8 m/s^2 * 2 s)
v_final = 4.9 m/s - 19.6 m/s
v_final = -14.7 m/s

The negative sign indicates that the rock is moving downward two seconds later. Therefore, the speed of the rock two seconds later is 14.7 m/s.

To find the rock's speed after two seconds, we need to calculate the magnitude of its velocity at that time.

Since the rock is thrown upwards, it will experience a constant acceleration due to gravity acting in the opposite direction. The acceleration due to gravity is approximately 9.8 m/s² (neglecting air resistance).

To find the rock's velocity after two seconds, we can use the kinematic equation:

v = u + at

Where:
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
t is the time elapsed.

In this case:
u = 4.9 m/s (initial velocity),
a = -9.8 m/s² (acceleration due to gravity),
t = 2 seconds (time elapsed).

Substituting the given values into the equation:

v = 4.9 m/s + (-9.8 m/s²) × 2 s

First, multiply the acceleration by the time:

v = 4.9 m/s - 19.6 m/s

Then, subtract the value:

v = -14.7 m/s

The negative sign indicates that the rock is moving in the opposite direction (downwards) compared to the positive initial velocity (upwards). Therefore, after two seconds, the rock has a speed of 14.7 m/s downwards.

Subtract g*t = 19.6 m/s^2 from +4.9 m/s.

The speed will be downward at that time.

Speed is a magnitude, and is always considered positive. So the answer is 14.7 m/s