ooo Sprint LTE 12:16 PM All Inboxes (1) A student stands at the edge of a cliff
ID: 1779264 • Letter: O
Question
ooo Sprint LTE 12:16 PM All Inboxes (1) A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18.0 m/s. The cliff is 50.0 m above a flat, horizontal beach. a. What are the coordinates of the initial position of the stone? b. What are the components of the initial velocity? c. Write the equations for the x- and y- components of the velocity of the stone with time. d. Write the equations for the position of the stone with time, using the coordinates. e. How long after being released does the stone strike the beach below the cliff? f. With what speed and angle of impact does the stone land?Explanation / Answer
a) The coordinates of the initial position of the stone
The stone is at the cliff which is at 50 m
So the y coordinate is 50 m, the x coordinate is 0 m
(xi, yi) = (0, 50m)
b) The initial velocity is
v 0 = 18 m/s
The ball is thrown horizontally, so there is only the horizontal component exist for the velocity
vx = 18 m/s
(vx , vy) = (18 m/s, 0)
c) Along the horizontal direction
Using Newton's equation of motion
v = u + a t
Where v is the velocity after the time t, u is the initial velocity and a is the acceleration
Along the horizontal direction
v = u (There is no acceleration along the x-direction)
The velocity in the horizontal direction is constant for all time
vx =18 m/s
Along the vertical direction
The acceleration is g in the downward direction
v = 0 - g t
vy = - g t
d) Using the Newton's equation
S = ut + (1/2) a t2
The horizontal distance with time
x = u t (acceleration is zero)
x = 18 t
The vertical distance with time
y = u t + (1/2) g t2
y = - (1/2) x 9.81 x t2
y = - 4.905 t2
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