Let's break down the problem step by step. ### **Given:** - Mass \( m = 0.9 \) kg - Initial velocity \( v_0 = 30 \) m/s (upward) - Height of building \( h_0 = 80 \) m - Gravity \( g = 9.8 \) m/s\(^2\) - Air resistance: \( F_{resist} = -m\frac{|v|}{30} \) (opposite to velocity) ### **Step 1: Equation of Motion** #### **For upward motion (until velocity = 0):** Forces: \[ ma = -mg - m\frac{v}{30} \] \[ a = -g - \frac{v}{30} \] This is a first-order linear differential equation: \[ \frac{dv}{dt} = -g - \frac{v}{30} \] #### **Step 2: Solve for Velocity as a Function of Time** \[ \frac{dv}{dt} + \frac{1}{30}v = -g \] This is a standard linear ODE. The integrating factor is \( e^{t/30} \): \[ e^{t/30} \frac{dv}{dt} + \frac{1}{30}e^{t/30} v = -g e^{t/30} \] \[ \frac{d}{dt} \left( e^{t/30} v \right) = -g e^{t/30} \] Integrate both sides: \[ e^{t/30} v = -g \int e^{t/30} dt + C \] \[ e^{t/30} v = -g \cdot 30 e^{t/30} + C \] \[ v = -30g + C e^{-t/30} \] Use initial condition \( v(0) = 30 \): \[ 30 = -30g + C \implies C = 30 + 30g \] So, \[ v(t) = -30g + (30 + 30g) e^{-t/30} \] --- ### **Step 3: Find Time to Reach Maximum Height** At max height, \( v = 0 \): \[ 0 = -30g + (30 + 30g) e^{-t/30} \] \[ 30g = (30 + 30g) e^{-t/30} \] \[ \frac{30g}{30 + 30g} = e^{-t/30} \] \[ \ln\left( \frac{30g}{30+30g} \right) = -\frac{t}{30} \] \[ t = -30 \ln\left( \frac{30g}{30+30g} \right) \] Plug in \( g = 9.8 \): \[ 30g = 294 \] \[ 30 + 30g = 30 + 294 = 324 \] \[ \frac{30g}{30+30g} = \frac{294}{324} = 0.9074 \] \[ t = -30 \ln(0.9074) \] \[ \ln(0.9074) = -0.0968 \] \[ t = -30 \times (-0.0968) = 2.904 \text{ s} \] So, **time to max height** \( t_{max} = \boxed{2.90} \) s. --- ### **Step 4: Find Maximum Height Above Ground** Now, integrate \( v(t) \) to get displacement from initial position. \[ v(t) = -30g + (30 + 30g) e^{-t/30} \] \[ h(t) = \int v(t) dt = -30g t - (30 + 30g) 30 e^{-t/30} + D \] \[ h(t) = -30g t - 30(30 + 30g) e^{-t/30} + D \] At \( t = 0, h(0) = 80 \): \[ 80 = 0 - 30(30 + 30g) \cdot 1 + D \] \[ 80 = -900 - 900g + D \] \[ D = 80 + 900 + 900g \] So, \[ h(t) = -30g t - 30(30 + 30g) e^{-t/30} + (80 + 900 + 900g) \] At \( t_{max} = 2.90 \) s: First, calculate \( e^{-2.90/30} = e^{-0.0967} = 0.908 \) Now, - \( 30g = 294 \) - \( 30 + 30g = 324 \) - \( 30 \times 324 = 9720 \) - \( 900g = 8820 \) - \( 80 + 900 = 980 \) - \( 980 + 8820 = 9800 \) \[ h(2.90) = -294 \times 2.90 - 9720 \times 0.908 + 9800 \] \[ -294 \times 2.90 = -852.6 \] \[ 9720 \times 0.908 = 8821.8 \] \[ h(2.90) = -852.6 - 8821.8 + 9800 = (9800 - 8821.8 - 852.6) \] \[ = 978.2 - 852.6 = 125.6 \] **Maximum Height Above Ground:** \[ \boxed{125.60 \text{ m}} \] --- ### **Step 5: Find Time When Ball Hits the Ground** Set \( h(t) = 0 \): \[ 0 = -294t - 9720 e^{-t/30} + 9800 \] \[ 294t + 9720 e^{-t/30} = 9800 \] \[ 9720 e^{-t/30} = 9800 - 294t \] \[ e^{-t/30} = \frac{9800 - 294t}{9720} \] \[ -t/30 = \ln\left(\frac{9800 - 294t}{9720}\right) \] \[ t = -30 \ln\left(\frac{9800 - 294t}{9720}\right) \] This is a **transcendental equation**; solve numerically. Let's try \( t \approx 8 \) s: - \( 294 \times 8 = 2352 \) - \( 9800 - 2352 = 7448 \) - \( 7448/9720 = 0.7664 \) - \( \ln(0.7664) = -0.266 \) - \( -30 \times -0.266 = 7.98 \), which is very close to our guess. **So, time is approximately \( t = 8.00 \) s.** --- ## **Final Answers** ### **a) Maximum height above the ground:** \[ \boxed{125.60 \ \text{m}} \] ### **b) Time when the ball hits the ground:** \[ \boxed{8.00 \ \text{seconds}} \]