Plsease include steps too Thank you!! A group of engineering students builds a m
ID: 1486031 • Letter: P
Question
Plsease include steps too Thank you!!
A group of engineering students builds a machine to launch a small package of emergency supplies across an 8.00 m wide canyon - to a group of stranded hikers, anxiously waiting on the other side. The students launch the package with an initial velocity of 12.0 m/s directed 60 decrees above the horizontal. At the instant of its launch, the package is located 4.00 m above the ground on the students' side of the canyon, at a horizontal distance of 2.00 m from the edge of students' side of the canyon, and 2.00 m above the plateau on the opposite side of the canyon, as shown in the Figure. Treat the package as a point particle (or a ball which has been thrown). Assume that air resistance is negligible, although it would not be. Assume that g = 9.80 m/S^2. How much time does it take the package to reach the highest point in its trajectory? At the instant when the package passes through the highest point in its trajectory, How much time would it take the package to travel a horizontal distance of 10.0 m from its launch point? At the instant when the package has traveled a horizontal distance of 10.0 m from its launch point, the scalar y component of the package's velocity would be The package would pass over the near edge of the plateau At the instant just before the package strikes the top surface of the plateau, the package's speed would be a True or False: The package lands on the plateau at a horizontal distance of ~ 3.8 m beyond the near side of theExplanation / Answer
The horizontal component of initial velocity is 12cos 60 = 6m/s
The vertical component of initial velocity is 12sin 60 = 10.392m/s
Horizontal distance covered is 8+2 = 10m from the launch position
Hence time taken = 10/6 = 5/3 = 1.67s
Now for vertical motion, a = -g, v = 10.392m/s, t = 1.67s
From 2nd equation of motion,
s = 10.392*1.67 - (1/2)(9.8)(1.67)2 = 3.7m
Hence, height at the near edge of the plateau is 3.7+2 = 5.7 above the top surface of plateau.
6) When the object hits the plateau:
Along vertical direction, s = -2m, a = -g, u = 10.392m/s
From 3rd equation of motion,
v2 = u2+2as
=> v2 = (10.392)2-2(9.8)(-2)
=> v = -12.1m/s (negative sign indicates downward direction)