A rock is thrown vertically upward from the ground with an initial speed 15m/s.

a. how high does go?
b. how much time is required for the rock to reach its maximum height?
c. what is the rock's height at t=2.00s?

At the top, final veloctiy is zero.

Vf=Vi-g*t to get time

h= Vi*t-1/2 g t^2 to get height.

To solve these problems, we can use the equations of motion. Let's break down each question step-by-step:

a. How high does it go?
To find the maximum height, we can use the formula for the height of an object in freefall:

h = (v₀²) / (2g)

Where:
h = height
v₀ = initial velocity (15 m/s)
g = acceleration due to gravity (9.8 m/s²)

Plugging in the values, we can calculate the height:

h = (15²) / (2 * 9.8)
h = 225 / 19.6
h ≈ 11.48 meters

Therefore, the rock goes up to approximately 11.48 meters.

b. How much time is required for the rock to reach its maximum height?
To find the time taken to reach the maximum height, we can use the formula:
t = v₀ / g

Plugging in the values:

t = 15 / 9.8
t ≈ 1.53 seconds

Thus, it takes approximately 1.53 seconds for the rock to reach its maximum height.

c. What is the rock's height at t=2.00 seconds?
To find the height at a specific time, we can use the formula for the height of an object in freefall:

h = v₀t - 0.5gt²

Plugging in the values:

h = (15 * 2) - (0.5 * 9.8 * (2²))
h = 30 - (0.5 * 9.8 * 4)
h = 30 - (0.5 * 39.2)
h = 30 - 19.6
h ≈ 10.4 meters

Therefore, at t=2.00 seconds, the rock's height is approximately 10.4 meters.

To answer these questions, we can use the equations of motion for free-falling objects under constant acceleration due to gravity. The key equation we will use is:

h = u*t + (1/2)*a*t^2

Where:
h is the height
u is the initial velocity
t is the time
a is the acceleration due to gravity (approximately -9.8 m/s^2)

Let's go through each question step by step.

a. How high does it go?
To find the maximum height, we need to determine when the rock reaches its highest point. At that point, its velocity will be zero. We can use the equation:

v = u + a*t

At maximum height, v = 0, so we can solve for t:

0 = 15 - 9.8*t

Solving for t, we find t = 1.53 seconds.

Substituting this value of t into the equation for height h, we get:

h = 15*1.53 + (1/2)*(-9.8)*(1.53)^2

Evaluating this expression, we find h ≈ 11.29 meters.

So, the rock reaches a height of approximately 11.29 meters.

b. How much time is required for the rock to reach its maximum height?
As calculated above, the time required for the rock to reach its maximum height is approximately 1.53 seconds.

c. What is the rock's height at t = 2.00 seconds?
To find the height at t = 2.00 seconds, we can again use the equation:

h = u*t + (1/2)*a*t^2

Substituting the given values:
u = 15 m/s (initial velocity)
t = 2.00 s

We have:

h = 15*2 + (1/2)*(-9.8)*(2)^2

Simplifying this expression, we find h ≈ 20.2 meters.

Therefore, the rock's height at t = 2.00 seconds is approximately 20.2 meters.