Let's solve the problem step by step. --- ## **Given Data** - Initial height, \( h_ = 2\,\text{m} \) - Initial velocity, \( v_ \) (to be found) - Launch angle, \( \alpha = 50^\circ \) - Wall height, \( H = 5\,\text{m} \) - Distance to wall, \( L \) (to be found in part b) - Acceleration due to gravity, \( g = 9.81\,\text{m/s}^2 \) --- ## **(a) Minimum \( v_ \) to clear the wall** ### **Step 1: Equations of Projectile Motion** - **Horizontal motion**: \( x(t) = v_ \cos\alpha \cdot t \) - **Vertical motion**: \( y(t) = h_ + v_ \sin\alpha \cdot t - \frac{1}{2}gt^2 \) At \( x = L \), \( y = H \): \[ L = v_ \cos\alpha \cdot t \] \[ H = h_ + v_ \sin\alpha \cdot t - \frac{1}{2}gt^2 \] Solve the first for \( t \): \[ t = \frac{L}{v_ \cos\alpha} \] Plug into the second: \[ H = h_ + v_ \sin\alpha \frac{L}{v_ \cos\alpha} - \frac{1}{2}g\left(\frac{L}{v_ \cos\alpha}\right)^2 \] Simplify: \[ H = h_ + L \tan\alpha - \frac{g L^2}{2 v_^2 \cos^2\alpha} \] ### **Step 2: Minimum \( v_ \) (occurs when projectile just clears wall)** Solve for \( v_ \): \[ H = h_ + L \tan\alpha - \frac{g L^2}{2 v_^2 \cos^2\alpha} \] \[ \frac{g L^2}{2 v_^2 \cos^2\alpha} = h_ + L \tan\alpha - H \] \[ v_^2 = \frac{g L^2}{2 \cos^2\alpha (h_ + L\tan\alpha - H)} \] **To minimize \( v_ \), the denominator must be maximized.** #### **Minimum \( v_ \) occurs when \( h_ + L\tan\alpha - H \) is maximized, i.e., for the smallest possible L.** But you are not given L or asked to find it yet. #### **Alternate approach:** The minimum \( v_ \) is when the projectile just passes over the wall. Set up for \( L \) as a function of \( v_ \), then minimize \( v_ \) for which the projectile just clears the wall. But since the question likely wants \( v_ \) as a function of L (since L is not given), we need to express \( v_ \) in terms of L and then find its minimum value. But usually, the minimum \( v_ \) occurs when the projectile passes directly over the wall at its peak (maximum height), but here \( h_ < H \), so the projectile must pass over at a point where \( y = H \), not at the peak. So, let's proceed: ### **Step 3: Express all in terms of L and solve for \( v_ \)** \[ v_^2 = \frac{g L^2}{2 \cos^2\alpha (h_ + L\tan\alpha - H)} \] The minimum happens when denominator is maximized, i.e. for the maximum possible value of \( L \) such that the projectile can still reach height \( H \). Alternatively, you can set up for the minimum \( v_ \) by considering the direct vertical throw and check, but that's not the case here since \( \alpha = 50^\circ \). So, **the minimum \( v_ \) is achieved at the smallest possible L for which the projectile can just pass over the wall at height H.** If L is not specified, express \( v_ \) in terms of L. --- ### **Final Answer for (a):** \[ \boxed{ v_ = \sqrt{ \frac{gL^2}{2\cos^2\alpha\,(h_ + L\tan\alpha - H)} } } \] Where: - \( g = 9.81\,\text{m/s}^2 \) - \( \alpha = 50^\circ \) - \( h_ = 2\,\text{m} \) - \( H = 5\,\text{m} \) --- ## **(b) For the value of \( v_ \) obtained in part (a), determine the distance L** Here, minimum \( v_ \) occurs when the projectile passes over the wall at the **peak** of its trajectory (i.e., maximum height). ### **Step 1: Maximum Height of the Projectile** \[ y_{max} = h_ + \frac{v_^2 \sin^2\alpha}{2g} \] Set \( y_{max} = H \): \[ H = h_ + \frac{v_^2 \sin^2\alpha}{2g} \] \[ \frac{v_^2 \sin^2\alpha}{2g} = H - h_ \] \[ v_^2 = \frac{2g (H - h_)}{\sin^2\alpha} \] \[ v_ = \sqrt{\frac{2g (H - h_)}{\sin^2\alpha}} \] Now, plug the numbers: - \( h_ = 2\,\text{m} \) - \( H = 5\,\text{m} \) - \( \alpha = 50^\circ \) - \( g = 9.81\,\text{m/s}^2 \) \[ v_ = \sqrt{\frac{2 \times 9.81 \times (5 - 2)}{\sin^2 50^\circ}} = \sqrt{\frac{2 \times 9.81 \times 3}{(\sin 50^\circ)^2}} \] \[ \sin 50^\circ \approx .766 \] \[ (\sin 50^\circ)^2 \approx .5868 \] \[ 2 \times 9.81 \times 3 = 58.86 \] \[ v_ = \sqrt{\frac{58.86}{.5868}} \approx \sqrt{100.39} \approx 10.02\,\text{m/s} \] **So, the minimum \( v_ \) is:** \[ \boxed{v_ = 10.\,\text{m/s}} \] --- ### **Step 2: Find the Distance \( L \) at which the projectile passes over the wall** At the peak, time to reach maximum height: \[ v_{y,\text{final}} = = v_ \sin\alpha - gt_{\text{peak}} \] \[ t_{\text{peak}} = \frac{v_ \sin\alpha}{g} \] Horizontal distance to peak: \[ L = v_ \cos\alpha \cdot t_{\text{peak}} \] \[ L = v_ \cos\alpha \cdot \left(\frac{v_ \sin\alpha}{g}\right) \] \[ L = \frac{v_^2 \sin\alpha \cos\alpha}{g} \] Plug in the numbers: - \( v_ = 10.02\,\text{m/s} \) - \( \sin 50^\circ = .766 \) - \( \cos 50^\circ = .6428 \) - \( g = 9.81\,\text{m/s}^2 \) \[ L = \frac{(10.02)^2 \times .766 \times .6428}{9.81} \] \[ = \frac{100.4 \times .4926}{9.81} \] \[ = \frac{49.47}{9.81} \approx 5.05\,\text{m} \] --- ### **Final Answers** ### (a) **Minimum initial velocity:** \[ \boxed{v_ = 10.\,\text{m/s}} \] ### (b) **Distance \( L \):** \[ \boxed{L = 5.1\,\text{m}} \] --- ### **Summary Table** | Part | Variable | Value | Units | |------|----------|-------|-------| | a | \( v_ \) (min) | 10. | m/s | | b | \( L \) | 5.1 | m | --- ## **Explanation** - We use projectile motion equations to relate the minimum velocity required to clear the wall at height \( H \). - The minimum occurs when the projectile passes over the wall at its peak. - We then use kinematics to find the corresponding distance \( L \) from the launch point to the wall. If you need further clarification or a symbolic answer for arbitrary values, let me know!