Question

A stone is thrown vertically upward with an initial speed of 8.00 m/s from the top of a 15.0 m tall building.Question 7Determine the time it took the stone to hit the groundRound your answer to 3 significant figures.Question 8Determine the speed of the stone the moment it hits the groundRound your answer to 3 significant figures.

Answer 1

Given,

The initial speed with which the stone was thrown upwards, u=8.00 m/s

The height of the building, h=15.0 m

The acceleration due to gravity is g=9.8 m/s²

From the equation of the motion we have,

`v=u+gt`

Where v is the final velocity of the stone and t is the time taken by the stone to reach the maximum height.

When the stone reaches the maximum height, the velocity of the stone becomes zero. And after that, the stone starts to fall back down.

Therefore the final velocity of the stone is, v=0 m/s. And in this case, the acceleration will be acting in the direction opposite to the direction of the motion of the stone. Thu the acceleration due to gravity will be, g=-9.8 m/s²

On substituting the known values in the equation,

`\begin{gathered} 0=8.00-9.8t \\ \Rightarrow t=(-8.00)/(-9.8) \\ t=0.82\text{ s} \end{gathered}`

The distance covered by the stone to travel to the maximum height is given by,

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

On substituting the known values,

`\begin{gathered} s=8.00*0.82+(1)/(2)*(-9.8)*0.82^2 \\ =3.26\text{ m} \end{gathered}`

Therefore the total distance the stone needs to cover, to hit the ground is,

`\begin{gathered} d=h+s \\ =15.0+3.26 \\ =18.26\text{ m} \end{gathered}`

Now we have to calculate the time duration in which the stone hits the ground.

The stone starts falling down from the maximum height, therefore the initial velocity of the stone, in this case, will be, u₀=0 m/s. And as we dont know the final velocity, we will need the distance traveled by the stone.

Therefore the time taken by the stone to hit the ground is given by the equation,

`s=u_0t_0+(1)/(2)gt^2_0`

Here g will have a positive value as the stone is now traveling downwards.

Therefore,

`\begin{gathered} 18.26=0+(1)/(2)*9.8* t^2_0 \\ t^2_0=(2*18.26)/(9.8) \\ =3.73\text{ } \\ \Rightarrow t_0=1.93\text{ s} \end{gathered}`

Therefore the time taken by the stone to hit the ground is,

`\begin{gathered} T=t_0+t \\ =1.93+0.82 \\ =2.75\text{ s} \end{gathered}`

Therefore the total time taken by the stone to hit the ground is 2.75 seconds

To determine the final velocity of the stone we use the equation,

`v_f=u_0+gt_0`

On substituting the known values,

`\begin{gathered} v_f=0+9.8*1.93 \\ =18.9\text{ m/s} \end{gathered}`

Therefore the final velocity of the stone when it hits the ground is 18.9 m/s