Answer:
y = 6.92[m]
Vf = 16.305[m/s]
Explanation:
To solve this problem we must use the following kinematics equations, where we must first find the final velocity.
[tex]v_{f}=v_{i}+(g*t)[/tex]
where:
Vi = initial velocity = 11.4[m/s]
Vf = final velocity [m/s]
t = time = 0.5[s]
g = gravity acceleration = 9.81[m/s^2}
Note: the positive sign in the equation above is because the acceleration of gravity goes in the direction of the motion of the rock.
Vf = 11.4 + (9.81*0.5)
Vf = 16.305 [m/s]
Now we can calculate how far the rock goes, using the following equation:
[tex]v_{f} ^{2}= v_{i} ^{2}+(2*g*y)\\v_{f} ^{2}- v_{i}^{2} = 2*g*y\\y = \frac{v_{f} ^{2}- v_{i}^{2}}{2*g} \\y = \frac{16.305^{2}-11.4^{2} }{2*9.81}\\ y=6.92[m][/tex]
Therefore the Rock will be (70 - 6.92) = 63.07 [m] above the water level
