A rock is thrown upward with a velocity of 27 meters per second from the top of a 39 meter high cliff, and it misses the cliff o

Question

A rock is thrown upward with a velocity of 27 meters per second from the top of a 39 meter high cliff, and it misses the cliff on the way back down. When will the rock be 9 meters from the water, below? Round your answer to two decimal places.

flash243 · Answer

This isnt too bad of simple kinematics physics question. For solving kinematics problem I like to use the four standard kinematics equations and the variables xi,xf,vi,vf,a,t which correspond to initial position, final position, initial velocity, final velocity, acceleration and time.
To solve this problem Ill separate it into two steps. First well find how far it went up and how long it took. To do this I know that
 
xi=0 m
xf=? m
vi=27 m/s
vf=0 m/s (b/c when it reaches its highest point it will have 0 velocity)
a = -9.8 m/s^2 (b/c gravity is always pulling down)
t = ?
 
Using the equation below:
vf^2= vi^2 + 2*a*(xf-xi) and plugging in the variables I can solve for xf
0^2 = 27^2 + 2*(-9.8)*(xf)
xf=37.9m
 
Now we solve for time:
vf = vi + a*t
0 = 27 + (-9.8)*t
t= 2.755 seconds
 
So now we know that it went 37 meters up and it took 2.755 seconds. If we add in the cliffs height that means it went for a total of 37.9+39 = 76.9 meters up. Now that we know how high it went up lets figure out how much time it took to go from 76.9 meters to 9 meters.
 
xi=0
xf=(76.9-9)=67.9 meters
vi=0
vf=?
a = 9.8 (positive this time b/c were moving in the same direction as gravity)
t = ?
 
Lets use a different kinematics equation this time
xf-xi=vi*t+(1/2)*a*t^2
67.9=0+(1/2)*(9.8)*t^2
Solving for t gives:
t=3.722
 
 
Now that we have our two times we can just add them to get an answer of 
t= 3.722+2.755 = 6.477

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A rock is thrown upward with a velocity of 27 meters per second from the top of a 39 meter high cliff, and it misses the cliff o

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