A person standing on the edge of a high cliff throws a rock straight up with an initial velocity of 1.09 m/s. The rock misses the edge of the cliff as it falls back to earth. Assume the rock lands at the bottom of the valley which is 7.34 m below the cliff, what is the speed(neglect the sign of the velocity) of the rock when it lands

Vf = 12.04 m/s

Explanation:

First, we consider the upward motion of the ball and use third equation of motion to find the height attained by the rock:

2gh' = Vf² - Vi²

where,

g = - 9.8 m/s² (negative sign for upward motion)

h' = height covered during upward motion = ?

Vf = Final Velocity = 0 m/s (since, rock stops at highest point)

Vi = Initial Velocity = 1.09 m/s

Therefore,

2(-9.8 m/s²)(h') = (0 m/s)² - (1.09 m/s)²

h' = (- 1.1881 m²/s²)/(- 19.6 m/s²)

h' = 0.06 m

Now, we analyze the downward motion of the rock. We use third equation of motion again:

2gh = Vf² - Vi²

where,

g = 9.8 m/s²

h = height covered during downward motion = 0.06 m + 7.34 m = 7.4 m

Vf = Final Velocity = ?

Vi = Initial Velocity = 0 m/s

Therefore,

2(9.8 m/s²)(7.4 m) = Vf² - (0 m/s)²

Vf = √(145.04 m²/s²)

Vf = 12.04 m/s

increasing transnational kinetic energy. because the transnational kinetic energy formula consists of only two variables, mass and velocity, increasing one of those properties is the only way to increase an object's transnational kinetic energy. increases to mass and velocity, however, do not have the same impact.

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