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

Answer:

(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.

Explanation:

Given that,

Time = 1.95 s

Distance = 23.5 m

(a). We need to calculate the velocity

Using equation of motion

$s = ut+\dfrac{1}{2}gt^2$

Put  the value into the formula

$-23.5=u\times1.95+\dfrac{1}{2}\times(-9.8)\times(1.95)^2$

$u=\dfrac{-23.5+4.9\times(1.95)^2}{1.95}$

$u=-2.496\ m/s$

(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

$-2.496=0-9.8\times t$

$t =\dfrac{2.496}{9.8}$

$t=0.254\ sec$

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

$s = ut+\dfrac{1}{2}gt^2$

Put the value into the formula

$s=0+\dfrac{1}{2}\times9.8\times(2.204)^2$

$s=23.80\ m$

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.