A rock is thrown at a window that is located 18.0 m above the ground. The rock is thrown at an angle of 40.0° above horizontal. The rock is thrown from a height of 2.00 m above the ground with a speed of 30.0 m/s and experiences no appreciable air resistance. If the rock strikes the window on its upward trajectory, from what horizontal distance from the window was it released?

Answer:27.32 mStep-by-step explanation:We are given that Height of window from the ground=18 mHeight of rock  form the ground=2 mSpeed of thrown rock=30 m/sWe have to find the horizontal distance from the window from which rock was release.Difference between window and rock=18-2=16 mInitial vertical velocity component=[tex]30sin40^{\circ}=19.28 [/tex]m/sInitial horizontal velocity component=[tex]30cos 40^{\circ}=22.98 m/s[/tex]If the rock is reached to maximum height Then, maximum height=[tex]\frac{v^2}{2g}=\frac{(19.28)^2}{2\times 9.8}=18.9731 m[/tex]Time taken by rock to reach maximum height=[tex]\frac{v}{g}=\frac{19.28}{9.8}=1.96775 s[/tex]Distance between window and maximum height at which rock reached=18.9731-16=2.973 mTime to drop 2.973 m=[tex]\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\cdot 2.973}{9.8}}=0.77893 s[/tex]Time to be at 16 m=1.96775-0.77893=1.189 sHorizontal distance=[tex]1.189\times 22.98=27.32 m[/tex]Hence, horizontal distance of rock from the window from which rock was released=27.32 m