Certainly! Let's first transcribe the problem from the image: --- **Question:** An object is thrown from a 60-m tall building with a vertical downward velocity of 5 m/s. - Acceleration due to gravity, \( g = 9.8 \text{ m/s}^2 \) - Assume \( y = \) is the ground, and upward is positive. - Your answer should be correct to two decimal places. **How many seconds will it take for the object to hit the ground?** --- Let's solve this step by step. --- ## Step 1: Write the Kinematic Equation The equation of motion for vertical displacement \( y \) is: \[ y = y_ + v_ t + \frac{1}{2} a t^2 \] Where: - \( y \) = final position (ground, so \( y = \)) - \( y_ \) = initial position (\( y_ = 60 \) m) - \( v_ \) = initial velocity (\( v_ = -5 \) m/s, since it's downward) - \( a \) = acceleration (\( a = -9.8 \) m/s², since acceleration is downward) - \( t \) = time in seconds --- ## Step 2: Substitute Values \[ = 60 + (-5)t + \frac{1}{2}(-9.8)t^2 \] \[ = 60 - 5t - 4.9t^2 \] --- ## Step 3: Rearrange the Equation \[ -4.9t^2 - 5t + 60 = \] Multiply both sides by \(-1\) (for clarity): \[ 4.9t^2 + 5t - 60 = \] --- ## Step 4: Solve the Quadratic Equation This is in the form \( at^2 + bt + c = \) with: - \( a = 4.9 \) - \( b = 5 \) - \( c = -60 \) The quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Plug in the values: ### Calculate Discriminant \[ b^2 - 4ac = (5)^2 - 4(4.9)(-60) \] \[ = 25 + 1176 \] \[ = 1201 \] ### Calculate \( t \) \[ t = \frac{-5 \pm \sqrt{1201}}{2 \times 4.9} \] \[ = \frac{-5 \pm 34.65}{9.8} \] We only consider the positive root (time can't be negative): \[ t = \frac{-5 + 34.65}{9.8} \] \[ = \frac{29.65}{9.8} \] \[ \approx 3.02 \] --- ## Final Answer \[ \boxed{3.02 \text{ s}} \] **It will take approximately 3.02 seconds for the object to hit the ground.**