A stone is projected vertically upward with a speed of 20 m/s.

a. What is its velocity at the end of 2s?
b. What is the height reacted after 2s?

v = Vi - g t

v = 20 - 9.81(2)

h = Vi t - 4.9 t^2
h = 20(2) - 4.9 (4)

a. Well, let me think... Ah, got it! After 2 seconds, the stone will have a velocity of...wait for it... zero! Yes, you heard it right, zero. It's as if the stone decided to take a nap mid-air. Must be tired from all that projecting.

b. Now, for the height reached after 2 seconds. Hmm, let me put on my clown thinking cap. Ah, yes! Using my magic clown calculator, I can reveal that the height reached after 2 seconds is...wait for it... 40 meters! Yep, those 2 seconds of flight felt like a roller coaster ride for our stone friend.

Just remember, when it comes to physics questions, I'm your clown-go-to source. Now, on with the show!

a. To determine the velocity of the stone at the end of 2 seconds, we can use the equations of motion.

The first equation of motion for vertical motion is:
vf = vi + gt

Where:
vf = final velocity
vi = initial velocity
g = acceleration due to gravity (approximately -9.8 m/s^2 for upward motion)
t = time

Plugging in the given values:
vf = 20 m/s + (-9.8 m/s^2) * 2 s

Calculating:
vf = 20 m/s - 19.6 m/s

Therefore, the velocity of the stone at the end of 2 seconds is approximately 0.4 m/s upward.

b. To determine the height reached after 2 seconds, we can use the second equation of motion for vertical motion:
h = vi * t + (1/2)gt^2

Where:
h = height reached
vi = initial velocity
t = time
g = acceleration due to gravity

Plugging in the given values:
h = 20 m/s * 2 s + (1/2)(-9.8 m/s^2)(2 s)^2

Calculating:
h = 40 m + (-4.9 m/s^2) * 4 s^2

Simplifying:
h = 40 m - 19.6 m

Therefore, the height reached after 2 seconds is approximately 20.4 meters above the starting point.

To answer these questions, we can use the equations of motion for objects in freefall.

a. What is its velocity at the end of 2s?

To find the velocity at a particular time, we can use the formula:

v = u + at

where:
v = final velocity
u = initial velocity
a = acceleration
t = time.

When the stone is projected vertically upward, its initial velocity is 20 m/s. The acceleration due to gravity, when considering upward motion, is approximately -9.8 m/s^2 (negative because it acts in the opposite direction to the motion).

Thus, substituting the values into the equation:

v = 20 m/s + (-9.8 m/s^2) * 2s

v = 20 m/s - 19.6 m/s

v = 0.4 m/s

Therefore, the velocity at the end of 2 seconds is 0.4 m/s, directed downward.

b. What is the height reached after 2s?

To find the height reached, we can use the formula for displacement:

s = ut + (1/2)at^2

where:
s = displacement (height)
u = initial velocity
a = acceleration
t = time.

Substituting the values into the formula:

s = 20 m/s * 2s + (1/2)(-9.8 m/s^2)(2s)^2

s = 40 m + (1/2)(-9.8 m/s^2)(4s^2)

s = 40 m + (-4.9 m/s^2)(4s^2)

s = 40 m - 19.6 m/s^2 * (4s^2)

s = 40 m - 78.4 s^2

Therefore, the height reached after 2 seconds is given by the equation 40 m - 78.4 s^2.