Question

A mountain lion decides to leap across a 5.00 m wide canyon, shown in the figure below. At the instant when the lion's feet lose contact with the ground, his velocity is 10.0 m/s directed 30.0 degrees above the horizontal. At this instant of "launch", the lion's center of mass is located 1.00 m above the ground, at a horizontal distance of 2.00 m from the near edge of the canyon, and 4.00 m above the level ground on the opposite side of the canyon, as shown in the figure below. In the following, treat the mountain lion as a point particle (located at his center of mass); that is, you can treat the mountain lion like a ball which has been pitched. Assume that air resistance is negligible, although it would not be so here. The lion (treated as a point particle) would pass over the opposite edge of the canyon about 1.4m above the level of the ground on the opposite side of the canyon. about 4.8 m above the level of the ground on the opposite side of the canyon. about 1.1 m above the level of the ground on the opposite side of the canyon. about 0.84 m above the level of the ground on the opposite side of the canyon The lion will strike the canyon wall about 0.84 m below the level of the ground on the opposite side. The lion (treated as a point particle) would land on the ground on the opposite side of the canyon about 13.4 m beyond the opposite edge of the canyon. about 8.4 m beyond the opposite edge of the canyon. The lion will fall short and land somewhere down in the canyon. about 15 m beyond the opposite edge of the canyon. about 6.4 m beyond the opposite edge of the canyon.

Answer 1

6) time taken to reach opposite edge = (5+2)/(10 cos 30 degree)

= 0.8082 s

vertical height =ut-0.5gt^2 =  5*0.8082 -4.9*0.8082^2

= 0.8 m

vertical height on opposide side = 0.8+4 = 4.8 m

option b is correct

7) using second equation of motion

- 4 = 5t - 4.9t^2

t = 1.548 s

excess distance =  10*cos 30 degree *1.548 - 7

= 6.4 m

option e is correct