A person is stranded and sitting 45m above the ground on the ledge of the cliff. of a safety rope is launched upward at an angle of 50 degrees above the horizontal at a distance of 60m away, what is the initial velocity of the rope so that it reaches the stranded person?
Y^2 = Yo^2 + 2g*h = 0 @ max ht.
Yo^2 = - 2(-9.8)*45 = 882
Yo = 29.7 m/s = Vertical component of
initial velocity.
Vo*sin50 = Yo = 29.7
Vo = 38.8 m/s.
To solve this problem, we can break it down into two separate components: horizontal and vertical.
Let's start with the horizontal component. We know that the rope is launched at an angle of 50 degrees above the horizontal and the distance to the person is 60m. We can use trigonometry, specifically the cosine function, to find the horizontal component of the initial velocity (Vx).
cos(θ) = adjacent / hypotenuse
In this case, adjacent is the horizontal component (Vx) and the hypotenuse is the initial velocity of the rope. Plugging in the given values:
cos(50) = Vx / initial velocity
Now, let's move on to the vertical component. We need to find the initial velocity in the y-direction (Vy). In this case, the person is sitting 45m above the ground, so we can use the equation of motion:
h = (Vyi x t) + (0.5 x g x t^2),
where h is the vertical displacement, Vyi is the initial velocity in the y-direction, t is the time of flight, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Given that the person is stranded and does not move vertically, the vertical displacement (h) is 0. Therefore:
0 = (Vy x t) + (0.5 x g x t^2)
Since it takes the same time for the rope to reach the person horizontally and vertically, we can substitute t/2 for t in the equation:
0 = (Vy x t/2) + (0.5 x g x (t/2)^2)
Now, we have two equations: one for the horizontal component (Vx) and one for the vertical component (Vy). We can solve these simultaneously to find the initial velocity of the rope (initial velocity):
cos(50) = Vx / initial velocity
0 = (Vy x t/2) + (0.5 x g x (t/2)^2)
Solve these equations simultaneously to find the values of Vx and Vy, then use the Pythagorean theorem to find the initial velocity:
initial velocity = sqrt(Vx^2 + Vy^2)