A student has fallen in a hole and must throw a key to their friend to go get a rope from the car. Which equation of motion will save the poor student (ground level is zero metres in height for throws of different speeds up from a height of negative two metres and time is in seconds)

Question options:

a)

h(t) = -4.9t² + 5.0t - 2.0

b)

h(t) = -4.9t² + 6.3t - 2.0

c)

h(t) = -4.9t² + 7.0t - 2.0

To determine which equation of motion will allow the key to reach the friend's hand, we need to find the equation where the key reaches the ground level (h = 0) at some positive time \( t \). Let's analyze each of the given equations by solving for \( t \) when \( h(t) = 0 \).

### Option (a):
\[ h(t) = -4.9t^2 + 5.0t - 2.0 \]

Set \( h(t) = 0 \):
\[ -4.9t^2 + 5.0t - 2.0 = 0 \]

This is a quadratic equation in the form \( at^2 + bt + c = 0 \). We can solve this using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = -4.9 \), \( b = 5.0 \), and \( c = -2.0 \):
\[ t = \frac{-5.0 \pm \sqrt{(5.0)^2 - 4(-4.9)(-2.0)}}{2(-4.9)} \]
\[ t = \frac{-5.0 \pm \sqrt{25.0 - 39.2}}{-9.8} \]
\[ t = \frac{-5.0 \pm \sqrt{-14.2}}{-9.8} \]

Since the discriminant (\( b^2 - 4ac \)) is negative, this equation has no real solutions. Hence, this option is not valid as the key cannot reach the ground level.

### Option (b):
\[ h(t) = -4.9t^2 + 6.3t - 2.0 \]

Set \( h(t) = 0 \):
\[ -4.9t^2 + 6.3t - 2.0 = 0 \]

Again, use the quadratic formula:
\[ t = \frac{-6.3 \pm \sqrt{(6.3)^2 - 4(-4.9)(-2.0)}}{2(-4.9)} \]
\[ t = \frac{-6.3 \pm \sqrt{39.69 - 39.2}}{-9.8} \]
\[ t = \frac{-6.3 \pm \sqrt{0.49}}{-9.8} \]
\[ t = \frac{-6.3 \pm 0.7}{-9.8} \]

This gives us two possible solutions:
\[ t = \frac{-6.3 + 0.7}{-9.8} = \frac{-5.6}{-9.8} = 0.571 \, \text{seconds} \]
\[ t = \frac{-6.3 - 0.7}{-9.8} = \frac{-7.0}{-9.8} = 0.714 \, \text{seconds} \]

Both values are positive and realistic. Thus, the key can reach the ground with this equation.

### Option (c):
\[ h(t) = -4.9t^2 + 7.0t - 2.0 \]

Set \( h(t) = 0 \):
\[ -4.9t^2 + 7.0t - 2.0 = 0 \]

Use the quadratic formula:
\[ t = \frac{-7.0 \pm \sqrt{(7.0)^2 - 4(-4.9)(-2.0)}}{2(-4.9)} \]
\[ t = \frac{-7.0 \pm \sqrt{49.0 - 39.2}}{-9.8} \]
\[ t = \frac{-7.0 \pm \sqrt{9.8}}{-9.8} \]
\[ t = \frac{-7.0 \pm 3.13}{-9.8} \]

This gives us two possible solutions:
\[ t = \frac{-7.0 + 3.13}{-9.8} = \frac{-3.87}{-9.8} = 0.395 \, \text{seconds} \]
\[ t = \frac{-7.0 - 3.13}{-9.8} = \frac{-10.13}{-9.8} = 1.034 \, \text{seconds} \]

Both values are positive and realistic. Thus, the key can reach the ground with this equation as well.

### Conclusion
Both options (b) and (c) are valid equations where the key reaches the ground. However, without more information about the desired timing or specific conditions, either of these could be considered correct.

Do you have any specific conditions or further questions on this problem?

### Related Questions

1. What is the maximum height reached by the key in each of these scenarios?
2. How does the initial velocity affect the time it takes for the key to reach the ground?
3. What is the significance of the discriminant in a quadratic equation related to projectile motion?
4. How can we determine the initial speed of the key from the given equations?
5. If the hole's depth was different, how would that affect the equations of motion?
6. How does gravity influence the motion of the key in these equations?
7. What other factors could influence the key's trajectory besides initial speed and gravity?
8. How can we use these equations to determine the speed of the key as it hits the ground?

### Tip

When solving quadratic equations, always check the discriminant (\( b^2 - 4ac \)) first to determine if there are real solutions, as it indicates whether the projectile reaches the desired height or not.

Learn which equation of motion will save a student who has fallen into a hole. Analyze quadratic equations for projectile motion and determine the correct equation for the key throw scenario. Suitable for high school students studying physics or algebra.

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