A person standing on the top of a cliff 171 ft high needs to throw a packet to his friend standing on the ground 228 ft horizontally away. If he throws the packet directly aiming at the friend with a speed of 15.0 ft/s , how short will the packet fall in ft?

Step-by-step solution

Let the point of projection be the origin (0,0) . The friend is on the ground at a horizontal distance of 228 ft and a vertical depth of 171 ft , so their coordinates are (228, -171) . Since the packet is aimed directly at the friend, the angle of projection below the horizontal is given by: = 171 228 = 3 4 This gives = 3 5 and = 4 5 . The initial speed of the packet is u = 15.0 ft/s . The horizontal and vertical components of the initial velocity are: u_x = u = 15 4 5 = 12 ft/s u_y = -u = -15 3 5 = -9 ft/s Taking