Question

A ball is dropped off a very tall canyon ledge.

Answer 1

Answer:

V = 49.05 [m/s]

Explanation:

We can easily find the result using kinematics equations, first, we will find the distance traveled during the 5 seconds.

$y =y_{o}+(v_{o}*t)+(\frac{1}{2}*g*t^{2} )$

where:

Yo = initial position = 0

y = final position [m]

Vo = initial velocity = 0

t = time = 5 [s]

g = gravity aceleration = 9.81 [m/s^2]

The initial speed is zero, as the body drops without imparting an initial speed. Therefore:

y = 0 + (0*5) + (0.5*9.81*5^2)

y = 122.625[m]

Now using the following equation we can find the speed it reaches during the 5 seconds.

$v_{f} ^{2}= v_{i} ^{2}+(2*g*y)\\v_{f}=\sqrt{2*9.81*122.625} \\v_{f}=49.05 [m/s]$