Question

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 1

Answer:

27.32 m

Step-by-step explanation:

We are given that

Height of window from the ground=18 m

Height of rock  form the ground=2 m

Speed of thrown rock=30 m/s

We have to find the horizontal distance from the window from which rock was release.

Difference between window and rock=18-2=16 m

Initial vertical velocity component=$30sin40^{\circ}=19.28$m/s

Initial horizontal velocity component=$30cos 40^{\circ}=22.98 m/s$

If the rock is reached to maximum height

Then, maximum height=$\frac{v^2}{2g}=\frac{(19.28)^2}{2\times 9.8}=18.9731 m$

Time taken by rock to reach maximum height=$\frac{v}{g}=\frac{19.28}{9.8}=1.96775 s$

Distance between window and maximum height at which rock reached=18.9731-16=2.973 m

Time to drop 2.973 m=$\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\cdot 2.973}{9.8}}=0.77893 s$

Time to be at 16 m=1.96775-0.77893=1.189 s

Horizontal distance=$1.189\times 22.98=27.32 m$

Hence, horizontal distance of rock from the window from which rock was released=27.32 m