Question

A certain freely falling object, released from rest, requires 1.95 s to travel the last 23.5 m before it hits the ground. (a) Find the velocity of the object when it is 23.5 m above the ground. (Indicate the direction with the sign of your answer. Let the positive direction be upward.) _______ m/s (b) Find the total distance the object travels during the fall. ________ m

Answer 1

(a). The velocity of the object is -2.496 m/s.

(b). The total distance of the object travels during the fall is 23.80 m.

Step-by-step explanation:

Given that,

Time = 1.95 s

Distance = 23.5 m

(a). We need to calculate the velocity

Using equation of motion

`s = ut+(1)/(2)gt^2`

Put the value into the formula

`-23.5=u*1.95+(1)/(2)*(-9.8)*(1.95)^2`

`u=(-23.5+4.9*(1.95)^2)/(1.95)`

`u=-2.496\ m/s`

(b). We need to calculate the total distance the object travels during the fall

Using equation of motion

`v = u+gt`

Put the value in the equation

`-2.496=0-9.8* t`

`t =(2.496)/(9.8)`

`t=0.254\ sec`

The total time is

`t'=t+1.95`

`t'=0.254+1.95`

`t'=2.204\ sec`

We need to calculate the distance

Using equation of motion

`s = ut+(1)/(2)gt^2`

Put the value into the formula

`s=0+(1)/(2)*9.8*(2.204)^2`

`s=23.80\ m`

Hence, (a). The velocity of the object is -2.496 m/s.

(b). The total distance of the object travels during the fall is 23.80 m.