An archer standing on a 15 degree slope shoots an arrow 20 degrees above the horizontal as shown in the attached figure. the arrow is shot with an initial speed of 50 m/s from a position 1.75m above the ground.
a) determine how far down the slope the arrow lands.
b) determine how long (in seconds) the arrow stays in the air.

This problem can easily be solved by rotating the x and y axes into x' and y' axes. Because of that, the gravity will have both x' and y' components. Let x' be parallel to the slope and in the same sense as the archer's arrow, and y' be perpendicular to the slope and pointing up. The components of gravity are g_y' = 9.8*sin(15), g_x' = 9.8*cos(15). Also, the arrow is shot at an angle 20+15=35 deg above the x' axis. 

The total time it takes for a projectile to stay in air is given by:

T = 2*v_y'/g_y' = 2*50*sin(35)/9.8sin(15) = 22.61 s.

The length traveled by the arrow along the slope is the distance it traveled along the x' axis. x' = 1/2*g_x'*T^2 + v_x'*T.

x' = 1/2*9.8cos(15)*22.61^2 + 50cos(35)*22.61x' = 3346 km.

I hope my answer has come to your help. Thank you for posting your question here in . We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

If i'm correct the answer is $15.97