<prompt> Etsa mohlala oa bothata ba fisiks bo kenyelletsang ho sisinyeha ha projectile.nan  <response>: Instruction in English: Create an example of a physics problem that involves projectile motion. . Response in English: A ball is thrown from the top of a cliff with an initial speed of 20 ms at an angle of 30 degrees above the horizontal. The cliff is 80 meters high. Determine the time it takes for the ball to reach the ground and the horizontal distance it travels during that time. To solve this problem, we can use the kinematics equations and projectile motion principles. First, we can split the initial velocity into its horizontal and vertical components. The horizontal component of the velocity is 20ms cos30 17.32 ms. The vertical component is 20ms sin30 10ms.Next, we can use the vertical motion to determine the time it takes for the ball to reach the ground. Using the kinematic equation y y0 v0y t 0.5 ay t2, where y0 is the initial height, v0y is the initial vertical velocity, ay is the acceleration in the y direction, and t is time, we can plug in the known values and solve for t. The acceleration in the y-direction is -9.8 ms2 due to gravity. Plugging in the known values, we have:0 80 10t - 4.9 t2Solving this quadratic equation we get t 3.67sNow, we can use the time and the horizontal component of the velocity to determine the horizontal distance traveled. The horizontal distance is simply the product of the horizontal speed and time traveled, which is 17.32 ms 3.67 s 63.5 m.Therefore, it takes the ball 3.67 seconds to reach the ground and it travels 63.5 meters horizontally. . Response in Sesotho: Ho lahleloa bolo ho tloha tlhōrōng ea lefika ka lebelo la pele la 20 ms ka lehlakoreng la likhato tse 30 ka holimo ho tse rapameng. Lefika le bophahamo ba limithara tse 80. Etsa qeto ea nako eo e e nkang hore bolo e fihle fatše le sebaka se tšekaletseng seo e se tsamaeang ka nako eo. Ho rarolla bothata bona, re ka sebelisa li-equation tsa kinematics le melao-motheo ea motsamao oa projectile. Taba ea pele, re ka arola lebelo la pele ka likarolo tsa eona tse tšekaletseng le tse otlolohileng. Karolo e otlolohileng ea lebelo ke 20ms cos30 17.32 ms. Karolo e emeng ke 20ms sin30 10ms. Ka mora moo, re ka sebelisa motsamao o otlolohileng ho bona hore na ho nka nako efe hore bolo e fihle fatše. Ho sebelisa equation ea kinematic y y0 v0y t 0.5 ay t2, moo y0 e leng bophahamo ba pele, v0y ke lebelo la pele le otlolohileng, ay ke ho potlakisa ka tsela ea y, me t ke nako, re ka khona. hokela makgabane a tsebahalang mme o rarolle bakeng sa t. Ho potlakisa tsela ea y ke -9.8 ms2 ka lebaka la matla a khoheli. Ho kenya litekanyetso tse tsejoang, re na le: 0 80 10t - 4.9 t2 Ho rarolla equation ena ea quadratic re fumana t 3.67s Joale, re ka sebelisa nako le karolo e tšekaletseng ea lebelo ho fumana sebaka se otlolohileng se tsamaileng. Sebaka se rapameng ke sehlahisoa sa lebelo le rapameng le nako e tsamaileng, e leng 17.32 ms 3.67 s 63.5 m. Ka hona, ho nka bolo metsotsoana e 3.67 ho fihla fatše me e tsamaea 63.5 limithara tse tšekaletseng. 