A rock is thrown upward with a velocity of 20 meters per second from the top of a 37 meter high cliff, and it misses the cliff on the way back down. When will the rock be 4 meters from the water, below? Round your answer to two decimal places.
Let us name
vr=velocity of the rock in the moment of throwing ( in m/s )
v0=velocity 0 m/s
vw=velocity of the rock 4 meters from the water.
hmax= top height of the rock measured from the water level
h= the height of the rock measured from the surface of the cliff.
hc= the height of the cliff
ha= the distance from the highest point the rock reaches by being thrown to the 4 meters depth of water.
Aplying Galilei's formula we have:
vr^2=2*g*h
so h=(vr^2)/2g=400/20=20m
But hmax=hc+h=37+20=57m
ha=57-(-4)=61m
So vw=sqrt(2*g*h)=34,93 m/s
To solve this problem, we can use the equations of motion to find the time at which the rock is 4 meters from the water below.
First, let's find the time it takes for the rock to reach its maximum height. We can use the equation:
v = u + at
where:
v = final velocity (0 m/s at the maximum height),
u = initial velocity (20 m/s),
a = acceleration due to gravity (-9.8 m/s²), and
t = time.
Rearranging the equation, we have:
t = (v - u) / a
Substituting the values, we get:
t = (0 - 20) / -9.8
t = 2.04 seconds (rounded to two decimal places).
Now, let's find the time it takes for the rock to fall from its maximum height to a point 4 meters from the water below. We can use the equation:
s = ut + 0.5at²
where:
s = distance (4 meters),
u = initial velocity (0 m/s at the maximum height),
a = acceleration due to gravity (-9.8 m/s²), and
t = time.
Rearranging the equation, we have:
t² = (2s) / a
Substituting the values, we get:
t² = (2 * 4) / -9.8
t² = -0.82
Taking the square root of both sides, we have:
t = √-0.82
Since time cannot be negative, the rock will never be 4 meters from the water below on its way back down.
Therefore, the rock will never be 4 meters from the water below.
To solve this problem, we can use the kinematic equation for the height of an object in free fall:
h = h0 + v0t - (1/2)gt^2
where:
h is the final height (4 meters below the cliff)
h0 is the initial height (37 meters above the water)
v0 is the initial velocity (20 meters per second, thrown upward)
g is the acceleration due to gravity (approximately 9.8 meters per second squared)
t is the time in seconds
Let's plug in the values and solve for t:
4 = 37 + 20t - (1/2)(9.8)t^2
To simplify the equation, let's multiply the second term by 2:
4 = 37 + 40t - 4.9t^2
Rearranging the equation to set it equal to zero:
-4.9t^2 + 40t + 4 = 0
Now we have a quadratic equation. We can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = -4.9, b = 40, and c = 4.
Plugging in the values:
t = (-40 ± √(40^2 - 4(-4.9)(4)))/(2(-4.9))
Calculating the expression inside the square root:
t = (-40 ± √(1600 + 78.4))/(2(-4.9))
t = (-40 ± √(1678.4))/(2(-4.9))
t = (-40 ± √(1678.4))/(-9.8)
Now we can calculate the two possible values for t:
t1 = (-40 + √(1678.4))/(-9.8)
t2 = (-40 - √(1678.4))/(-9.8)
Evaluating t1:
t1 = (-40 + √(1678.4))/(-9.8)
t1 = (-40 + 40.96)/(-9.8)
t1 = 0.96/(-9.8)
t1 ≈ -0.098
Since time cannot be negative, we disregard t1.
Now evaluating t2:
t2 = (-40 - √(1678.4))/(-9.8)
t2 = (-40 - 40.96)/(-9.8)
t2 = -80.96/(-9.8)
t2 ≈ 8.27
Therefore, the rock will be 4 meters from the water below approximately 8.27 seconds after it was thrown upward.