To solve the problem we will simply perform equivalence between both expressions. We will proceed to place your units and develop your internal operations in case there is any. From there we will compare and look at its consistency
At the same time we have that
Therefore there is not have same units and both are not consistent and the correct answer is B.
Answer:
The rock must leave the cliff at a velocity of 28.2 m/s
Explanation:
The position vector of the rock at a time t can be calculated using the following equation:
r = (x0 + v0x · t, y0 + 1/2 · g · t²)
Where:
r = position vector at time t.
x0 = initial horizontal position.
v0x = initial horizontal velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).
Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the cliff so that x0 and y0 = 0.
When the rock reaches the ground, the position vector will be (see r1 in the figure):
r1 = (90 m, -50 m)
Then, using the equation of the vector position written above:
90 m = x0 + v0x · t
-50 m = y0 + 1/2 · g · t²
Since x0 and y0 = 0:
90 m = v0x · t
-50 m = 1/2 · g · t²
Let´s use the equation of the y-component of the vector r1 to find the time it takes the rock to reach the ground and with that time we can calculate v0x:
-50 m = 1/2 · g · t²
-50 m = -1/2 · 9.81 m/s² · t²
-50 m / -1/2 · 9.81 m/s² = t²
t = 3.19 s
Now, using the equation of the x-component of r1:
90 m = v0x · t
90 m = v0x · 3.19 s
v0x = 90 m / 3.19 s
v0x = 28.2 m/s
Answer:
Explanation:
Given
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