Question

An object is dropped from a height of 65 m above ground level. A) determine the final speed in m/s, at which the object hits the ground c) determine the distance in meters, traveled during the last second of motion before hitting the ground.

Answer 1

Given:

height = 65 m

Given that the object is in free fall, let's solve for the following:

• (a). determine the final speed in m/s.

To find the final velocity, apply the kinematics equation:

`v^2=u^2-2ax`

Where:

v is the final velocity

u is the initial velocity = 0

a is the acceleration due to gravity = 9.8 m/s²

x is the displacement = 65 m

Thus, we have:

`\begin{gathered} v^2=0^2-2(-9.8)(65) \\ \\ v^2=-(-1274) \\ \\ v^2=1274 \\ \\ \text{ Take the square root of both sides:} \\ √(v^2)=√(1274) \\ \\ v=35.69\text{ m/s} \end{gathered}`

Therefore the final speed will be -35.69 m/s.

• (c). The distance traveled during the last second of motion before hitting the ground.

To find the distance, apply the formula:

`H=ut+(1)/(2)at^2`

Where:

H is the height.

u is the initial velocity = 0 m/s

t is the time

a is acceleration due to gravity.

Let's rewrite the formula to find the time traveled.

`\begin{gathered} H=0t+(1)/(2)at^2 \\ \\ H=(1)/(2)at^2 \\ \\ t=\sqrt{(2H)/(a)} \end{gathered}`

Thus, we have:

`\begin{gathered} t=\sqrt{(2*65)/(9.8)} \\ \\ t=\sqrt{(130)/(9.8)} \\ \\ t=√(13.26) \\ \\ t=3.64\text{ s} \end{gathered}`

Therefore, the time is 3.64 seconds.

Now, to find the distance traveled during the last second of motion, apply the formula:

`s=(1)/(2)a(t_2^2-t_1^2)`

Where:

t2 = 3.64 seconds

t1 = 3.64 seconds - 1 second = 2.64 seconds

Thus, we have:

`\begin{gathered} s=(1)/(2)(9.8)((3.64)^2-(2.64)^2) \\ \\ s=4.9(13.2496-6.9696) \\ \\ s=4.9(6.28) \\ \\ s=30.77 \end{gathered}`

Therefore, the distance in meters, traveled during the last second of motion before hitting the ground is 30.77 meters.

ANSWER:

(A). -35.69 m/s

(C). 30.77 m