(a) A man stands on the roof of a building 50 m high and throws stone with a speed of \( 60 \mathrm{~m} / \mathrm{s} \) at an angle of \( 37^{\circ} \) above the horizontal. Calculate (I) the maximum height above the roof reached by the rock (ii) the magnitude of the downward velocity of the rock just before it strikes the ground. (iii) the horizontal distance from the base of the building to the point where the stone strikes the ground.

To solve the problem, we will break it down into three parts as requested. ### Given Data: - Height of the building, \( h = 50 \, \text{m} \) - Initial speed of the stone, \( v_0 = 60 \, \text{m/s} \) - Angle of projection, \( \theta = 37^\circ \) ### Step 1: Resolve the initial velocity into horizontal and vertical components. The horizontal component of the initial velocity \( v_{0x} \) is given by: \[ v_{0x} = v_0 \cos(\theta) = 60 \cos(37^\circ) \] Using \( \cos(37^\circ) \approx 0.8 \): \[ v_{0x} = 60 \times 0.8 = 48 \, \text{m/s} \] The vertical component of the initial velocity \( v_{0y} \) is given by: \[ v_{0y} = v_0 \sin(\theta) = 60 \sin(37^\circ) \] Using \( \sin(37^\circ) \approx 0.6 \): \[ v_{0y} = 60 \times 0.6 = 36 \, \text{m/s} \] ### Part (I): Maximum height above the roof reached by the rock The maximum height \( H \) above the roof can be calculated using the formula: \[ H = \frac{v_{0y}^2}{2g} \] where \( g \approx 9.81 \, \text{m/s}^2 \). Substituting the values: \[ H = \frac{(36)^2}{2 \times 9.81} = \frac{1296}{19.62} \approx 66.05 \, \text{m} \] Thus, the maximum height above the roof is approximately \( 66.05 \, \text{m} \). ### Part (II): Magnitude of the downward velocity just before it strikes the ground To find the downward velocity just before it strikes the ground, we can use the kinematic equation: \[ v_y^2 = v_{0y}^2 + 2g(h + H) \] where \( h + H \) is the total height from which the stone falls. Calculating \( h + H \): \[ h + H = 50 + 66.05 = 116.05 \, \text{m} \] Now substituting the values: \[ v_y^2 = (36)^2 + 2 \times 9.81 \times 116.05 \] Calculating: \[ v_y^2 = 1296 + 2275.81 \approx 3571.81 \] Taking the square root: \[ v_y \approx \sqrt{3571.81} \approx 59.8 \, \text{m/s} \] Thus, the magnitude of the downward velocity just before it strikes the ground is approximately \( 59.8 \, \text{m/s} \). ### Part (III): Horizontal distance from the base of the building to the point where the stone strikes the ground To find the time of flight \( t \), we can use the vertical motion equation: \[ y = v_{0y} t - \frac{1}{2} g t^2 \] Setting \( y = -50 \, \text{m} \) (since it falls 50 m below the roof): \[ -50 = 36t - \frac{1}{2} \times 9.81 t^2 \] Rearranging gives: \[ 4.905t^2 - 36t - 50 = 0 \] Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): - \( a = 4.905 \) - \( b = -36 \) - \( c = -50 \) Calculating the discriminant: \[ D = (-36)^2 - 4 \times 4.905 \times (-50) = 1296 + 980.4 = 2276.4 \] Now calculating \( t \): \[ t = \frac{36 \pm \sqrt{2276.4}}{2 \times 4.905} \] Calculating \( \sqrt{2276.4} \approx 47.7 \): \[ t = \frac{36 \pm 47.7}{9.81} \] Taking the positive root: \[ t = \frac{83.7}{9.81} \approx 8.54 \, \text{s} \] Now, the horizontal distance \( d \) is given by: \[ d = v_{0x} t = 48 \times 8.54 \approx 409.92 \, \text{m} \] Thus, the horizontal distance from the base of the building to the point where the stone strikes the ground is approximately \( 409.92 \, \text{m} \). ### Summary of Results: 1. Maximum height above the roof: \( 66.05 \, \text{m} \) 2. Downward velocity just before striking the ground: \( 59.8 \, \text{m/s} \) 3. Horizontal distance from the base of the building: \( 409.92 \, \text{m} \)