A rock is thrown vertically upwards by a hand which is 1.2 meters above the ground. The displacement from the instant it leaves the hand is given by the function
s(t)=-4.9t^2+10t
	Find the maximum displacement from the point where the rock left the hand.

	Find the time taken from the rock being released by the hand until it reaches the ground again.
c)	Find the velocity of the rock at the instance it hits the ground again.

Let's solve each part of the problem one by one.

### a) Maximum Displacement from the Point Where the Rock Left the Hand

The displacement function is given by:

\[
s(t) = -4.9t^2 + 10t
\]

To find the maximum displacement, we need to find the vertex of this quadratic function. The vertex of a parabola described by \( ax^2 + bx + c \) is at:

\[
t = -\frac{b}{2a}
\]

Here, \( a = -4.9 \) and \( b = 10 \). Plugging these values in:

\[
t = -\frac{10}{2(-4.9)} = \frac{10}{9.8} = 1.02 \, \text{seconds}
\]

Now, we substitute \( t = 1.02 \) back into the displacement function to find the maximum displacement:

\[
s(1.02) = -4.9(1.02)^2 + 10(1.02)
\]

Calculating:

\[
s(1.02) = -4.9(1.0404) + 10.2 = -5.098 + 10.2 = 5.102 \, \text{meters}
\]

So, the maximum displacement from the point where the rock left the hand is approximately **5.1 meters**.

### b) Time Taken from Release Until the Rock Reaches the Ground Again

The rock reaches the ground when the displacement \( s(t) \) is zero. We set the displacement function to zero and solve for \( t \):

\[
0 = -4.9t^2 + 10t
\]

Factor the quadratic equation:

\[
0 = t(-4.9t + 10)
\]

This gives us two solutions:

\[
t = 0 \quad \text{or} \quad t = \frac{10}{4.9} \approx 2.04 \, \text{seconds}
\]

Since \( t = 0 \) is the time when the rock was released, the time taken until it reaches the ground again is **2.04 seconds**.

### c) Velocity of the Rock at the Instant It Hits the Ground

The velocity function is the derivative of the displacement function \( s(t) \):

\[
v(t) = \frac{ds(t)}{dt} = -9.8t + 10
\]

We want to find the velocity at the time when the rock hits the ground, \( t = 2.04 \) seconds:

\[
v(2.04) = -9.8(2.04) + 10
\]

Calculating:

\[
v(2.04) = -19.992 + 10 = -9.992 \, \text{m/s}
\]

So, the velocity of the rock when it hits the ground is approximately **-9.99 m/s** (the negative sign indicates that the rock is moving downward).

### Summary:
- **Maximum displacement from the release point:** 5.1 meters
- **Time taken to reach the ground:** 2.04 seconds
- **Velocity upon hitting the ground:** -9.99 m/s

Would you like a more detailed explanation, or do you have any questions?

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**Related Questions:**
1. How would the problem change if the initial height was different?
2. What is the significance of the negative sign in the velocity upon hitting the ground?
3. Can you find the initial velocity of the rock when it leaves the hand?
4. What would be the maximum height above the ground if the hand was at 1.5 meters instead?
5. How does the gravitational constant affect the time the rock stays in the air?

**Tip:** When working with quadratic functions, identifying the vertex can quickly give you the maximum or minimum value of the function.

A rock is thrown vertically upwards by a hand which is 1.2 meters above the ground. The displacement from the instant it leaves the hand is given by the function s(t) = -4.9t^2 + 10t. Find the maximum displacement from the point where the rock left the hand. Find the time taken from the rock being released by the hand until it reaches the ground again. Find the velocity of the rock at the instance it hits the ground again.

Explore the vertical motion of a rock with a displacement function s(t) = -4.9t^2 + 10t. Find the maximum displacement, the time taken for the rock to return to the ground, and the velocity upon impact. This problem covers kinematics, quadratic functions, and calculus concepts, ideal for Grades 11-12 students.

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