A student stands at the edge of a cliff and throws a stonehorizontally over the
ID: 1666872 • Letter: A
Question
A student stands at the edge of a cliff and throws a stonehorizontally over the edge with a speed of 27 m/s. The cliff is 53 mabove a flat, horizontal beach as shown in the figure below. What are the coordinates of the initial position of thestone? What are the components of the initial velocity? Write the equations for the velocity of the stone with time.(Use g for the acceleration of gravity, and t as necessary.) Write the equations for the position of the stone with time.(Use g for the acceleration of gravity, and t as necessary.) How long after being released does the stone strike the beachbelow the cliff?With what speed and angle of impact does the stone land? A student stands at the edge of a cliff and throws a stonehorizontally over the edge with a speed of 27 m/s. The cliff is 53 mabove a flat, horizontal beach as shown in the figure below. What are the coordinates of the initial position of thestone? What are the components of the initial velocity? Write the equations for the velocity of the stone with time.(Use g for the acceleration of gravity, and t as necessary.) Write the equations for the position of the stone with time.(Use g for the acceleration of gravity, and t as necessary.) How long after being released does the stone strike the beachbelow the cliff?
With what speed and angle of impact does the stone land?
Explanation / Answer
I would set the initial coordinates as X = 0 , Y = 53m
What are the components of the initial velocity?
The initial velocity is broken into the Vertical and HorizontalMotions
Vx = 27m/s
Vy = 0 m/s
Write the equations for the velocity of the stone with time.(Use g for the acceleration of gravity, and t as necessary.)
Again, these need to be broken into the Vertical and HorizontalComponents
Vy = -g*t+Vy (Velocity in y direction is neg gravity timestime, plus initial velocity)
Vx = Sx / t (I am assuming nodeceleration in X direction, Sx = Position in X direction)
Write the equations for the position of the stone with time.(Use g for the acceleration of gravity, and t as necessary.)
Y(t) = (1/2)*g*t^2 + V0y*t + H0 (V0y = initial velocity inY direction H0 = Initial position in Y Direction)
X(t) = (Vx * t) + X0
How long after being released does the stone strike the beachbelow the cliff?
The stone will stop when the Y(t) = 0
Use Equation t = ((2*dy)/g)^(1/2) (Here dy = 53meters and g = 9.81 m/s)
t= 3.28 seconds
With what speed and angle of impact does the stone land?
Speed is V = (Vy^2+Vx^2)^(1/2)
Vy = (2*g*h)^(1/2) (h = 53 metershere)
Vy = 32.24 m/s
Vx = 27 m/s (This was given)
To Solve for the angle you should use
Tan(theta)=Vy/Vx
Theta = ArcTan(Vy/Vx)
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