Question

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

Answer 1

Answer:

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