A 12.0-kg shell is launched at an angle of 55.0o above the

Solution 110P Step 1: W can resolve the components of velocity as vertical and horizontal. Vertical component of velocity, v verticalv sin Provided, it is making an angle, = 55° with the horizontal Then, v vertical150 m/s × sin 55 = 150 m/s× 0.8192 = 122.87 m/s Since it is moving against the gravitational force, we can write, v - u = - 2 gS Where, v - final velocity, u - initial velocity, g - acceleration due to gravity And S - distance travelled In vertical motion, v = 0 m/s at a point 2 2 2 2 Therefore, 0 - 122.87 m /s = - 2 × 9.8 m/s × S The vertical displacement, S = 15098 m /s / 19.6 m/s = 770.26 m 2 Similarly, v = u - gt 0 = 122.87 m/s - 9.8 t Rearranging, t = 122.87 m/s / 9.8 m/s = 12.54 s 2 The time to reach maximum height will be, t = 12.54 s Therefore, the horizontal distance for explosion, H = v horizontal vhorizontal cos Provided, it is making an angle, = 55° with the horizontal Then, v horizontal50 m/s × cos 55 = 150 m/s× 0.5736 = 86.04 m/s Therefore, the horizontal distance for explosion, H = 86.04 m/s×12.54 s = 1078.9 m