an object is ejected sideways with a speed of 10 meters per second, a height of 78.4 meters above the ground. How long does it take to reach the ground?
To determine how long it takes for the object to reach the ground, we can use the kinematic equation for vertical motion:
\[ s = ut + \frac{1}{2}gt^2\]
where:
- \(s\) is the vertical displacement or height of the object above the ground (\(78.4\) meters in this case),
- \(u\) is the initial vertical velocity of the object (0 m/s since the object is ejected sideways),
- \(t\) is the time taken for the object to reach the ground (what we want to find), and
- \(g\) is the acceleration due to gravity (-9.8 m/s²).
Since the object is ejected sideways with an initial velocity of 10 m/s, it means the object has no initial vertical velocity. Therefore, the equation becomes:
\[ s = \frac{1}{2}gt^2\]
Substituting the given values, we have:
\[ 78.4 = \frac{1}{2} \times (-9.8) \times t^2\]
Simplifying the equation further:
\[ 78.4 = -4.9t^2\]
To isolate \(t^2\), divide both sides by -4.9:
\[ \frac{78.4}{-4.9} = t^2\]
\[ t^2 = -16\]
Since time cannot be negative in this context, there is no real solution for \(t^2 = -16\).
Therefore, the given combination of the object being ejected sideways and its height above the ground does not allow us to calculate the time it takes for the object to reach the ground using the given information.