Questions: A stone thrown downward with an initial velocity of 39.2 m / sec will travel a distance of s meters, where s(t)=4.9 t^2+39.2 t and t is in seconds. If a stone is thrown downward at 39.2 m / sec from a height of 411.6 m, how long will it take the stone to hit the ground? It will take seconds to hit the ground.

Solution Steps

The distance \( s(t) \) traveled by the stone is given by the equation: \[ s(t) = 4.9t^2 + 39.2t \]

The stone is thrown from a height of 411.6 meters, so we need to find the time \( t \) when the stone has traveled 411.6 meters downward. Therefore, we set \( s(t) \) equal to 411.6 meters: \[ 4.9t^2 + 39.2t = 411.6 \]

Rearrange the equation to standard quadratic form: \[ 4.9t^2 + 39.2t - 411.6 = 0 \]

We solve this quadratic equation using the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 4.9 \), \( b = 39.2 \), and \( c = -411.6 \).

First, calculate the discriminant: \[ \Delta = b^2 - 4ac = (39.2)^2 - 4(4.9)(-411.6) \] \[ \Delta = 1536.64 + 8056.32 = 9592.96 \]

Now, substitute the values into the quadratic formula: \[ t = \frac{-39.2 \pm \sqrt{9592.96}}{2 \times 4.9} \] \[ t = \frac{-39.2 \pm 97.94}{9.8} \]

We have two potential solutions: \[ t_1 = \frac{-39.2 + 97.94}{9.8} \approx 6 \] \[ t_2 = \frac{-39.2 - 97.94}{9.8} \approx -14 \]

Since time cannot be negative, we discard \( t_2 \).

Final Answer