Explanation:
To find the time it takes for the object to reach the ground, we can use the kinematic equation for motion under constant acceleration (gravity). The equation is:
\[ h = v_i t + \frac{1}{2} a t^2 \]
where:
- \( h \) is the height (60 m)
- \( v_i \) is the initial velocity (10 m/s)
- \( a \) is the acceleration due to gravity (9.8 m/s²)
- \( t \) is the time
We need to solve for \( t \). Plugging in the values:
\[ 60 = 10t + \frac{1}{2} \cdot 9.8 \cdot t^2 \]
This simplifies to:
\[ 60 = 10t + 4.9t^2 \]
Rearranging the equation into standard quadratic form:
\[ 4.9t^2 + 10t - 60 = 0 \]
We can solve this quadratic equation using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 4.9 \), \( b = 10 \), and \( c = -60 \). Plugging in these values:
\[ t = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 4.9 \cdot (-60)}}{2 \cdot 4.9} \]
\[ t = \frac{-10 \pm \sqrt{100 + 1176}}{9.8} \]
\[ t = \frac{-10 \pm \sqrt{1276}}{9.8} \]
\[ t = \frac{-10 \pm 35.7}{9.8} \]
This gives us two solutions:
\[ t = \frac{25.7}{9.8} \approx 2.62 \, \text{seconds} \]
and
\[ t = \frac{-45.7}{9.8} \approx -4.66 \, \text{seconds} \] (which is not physically meaningful in this context)
Thus, the time it takes for the object to reach the ground is approximately 2.62 seconds.
