Question

A ball is dropped off a very tall canyon ledge.

How fast is the ball traveling after 5
seconds?

Answer 1

V = 49.05 [m/s]

Step-by-step 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)+((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)=√(2*9.81*122.625) \\v_(f)=49.05 [m/s]`