Question

A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. The stone is launched 50.0 m above the ground, and the stone just misses the edge of the roof on the way down.

1. Using t=0 s as the time the stone leaves the thrower's hand at position, determine the time at which the stone reaches its maximum height.

2. Find the maximum height of the stone

3. Determine the velocity of the stone when it returns to the height from which it was thrown

4. Find the position of the stone at t=5.00 s

Answer 1

Concepts and reason

The concept used in this problem is Newton’s equations of motion. The time taken by stone, the maximum height, the velocity and the position can be determined using equations of motion.

Fundamentals

Newton’s equations of motion:

The Newton’s equations of motion give the relation between initial speed, final speed, acceleration, distance and time of a motion. These equations describe the motion of an object.

Three equations of motion are:

$v - u = at$ …… (1)

$s = ut + \frac{1}{2}a{t^2}$ …… (2)

${v^2} - {u^2} = 2as$ …… (3)

Here, $v$ is the final speed, $u$ is the initial speed, $a$ is the acceleration, $s$ is the distance and $t$ is the time.

(1)

The first equation of motion is,

$v - u = at$

Rearranging the above equation,

$t = \frac{{v - u}}{a}$

At maximum height, the final speed of the stone is zero.

Substitute $0{\rm{ m/s}}$ for $v$ , $20{\rm{ m/s}}$ for $u$ and $- 9.8{\rm{ m/}}{{\rm{s}}^2}$ for $a$ in the above equation.

$\begin{array}{c}\\t = \frac{{0{\rm{ m/s}} - 20{\rm{ m/s}}}}{{ - {\rm{9}}{\rm{.8 m/}}{{\rm{s}}^2}}}\\\\ = 2.04{\rm{ s}}\\\end{array}$

(2)

The second equation of motion is,

$s = ut + \frac{1}{2}a{t^2}$

Here, $s$ is the maximum height of the stone.

Substitute $2.04{\rm{ s}}$ for $t$ , $20{\rm{ m/s}}$ for $u$ and $- 9.8{\rm{ m/}}{{\rm{s}}^2}$ for $a$ .

$\begin{array}{c}\\s = \left( {20{\rm{ m/s}}} \right)\left( {2.04{\rm{ s}}} \right) + \frac{1}{2}\left( { - 9.8{\rm{ m/}}{{\rm{s}}^2}} \right){\left( {2.04{\rm{ s}}} \right)^2}\\\\ = 40.8{\rm{ m}} - 20.4{\rm{ m}}\\\\ = 20.4{\rm{ m}}\\\end{array}$

(3)

The first equation of motion is,

$v' - u = at'$

$v' = u + at'$ …… (4)

Here, $t'$ is the time taken by stone to come at the position it was thrown and $v'$ is the velocity of the stone at that position.

Time taken by stone to come at the height it was thrown from the maximum height is,

$t' = 2t$

Substitute $2.04{\rm{ s}}$ for $t$ in the above equation.

$\begin{array}{c}\\t' = 2\left( {2.04{\rm{ s}}} \right)\\\\ = 4.08{\rm{ s}}\\\end{array}$

Substitute $20{\rm{ m/s}}$ for $u$ , $4.08{\rm{ s}}$ for $t'$ and $- 9.8{\rm{ m/}}{{\rm{s}}^2}$ for $a$ in the equation (4).

$\begin{array}{c}\\v' = \left( {20{\rm{ m/s}}} \right) + \left( { - 9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {4.08{\rm{ s}}} \right)\\\\ = - 20{\rm{ m/s}}\\\end{array}$

The negative sign indicates the downward motion of the stone. The velocity is,

$v = 20{\rm{ m/s}}$

(4)

The second equation of motion is,

${s_1} = u{t_1} + \frac{1}{2}a{t_1}^2$

Here, ${s_1}$ is the position at given time and ${t_1}$ is the given time.

Substitute $5{\rm{ s}}$ for ${t_1}$ , $20{\rm{ m/s}}$ for $u$ and $- 9.8{\rm{ m/}}{{\rm{s}}^2}$ for $a$ in the above equation.

$\begin{array}{c}\\{s_1} = \left( {20{\rm{ m/s}}} \right)\left( {5{\rm{ s}}} \right) + \frac{1}{2}\left( { - 9.8{\rm{ m/}}{{\rm{s}}^2}} \right){\left( {5{\rm{ s}}} \right)^2}\\\\ = 100{\rm{ m}} - 122.5{\rm{ m}}\\\\ = - 22.5{\rm{ m}}\\\end{array}$

The position of stone from the ground at given time is,

$x = 50{\rm{ m}} + {s_1}$

Substitute $- 22.5{\rm{ m}}$ for ${s_1}$ in the above equation.

$\begin{array}{c}\\x = 50{\rm{ m}} - 22.5{\rm{ m}}\\\\ = 27.5{\rm{ m}}\\\end{array}$

Ans: Part 1

The time taken by stone in reaching the maximum height is ${\bf{ 2}}{\bf{.04 s}}$ .

Part 2

The maximum height of the stone is ${\bf{ 20}}{\bf{.4 m}}$ .

Part 3

The return velocity of the stone is ${\bf{ 20 m/s}}$ .

Part 4

The position of the stone at ${\bf{5 s}}$ is ${\rm{ }}{\bf{27}}{\bf{.5 m}}$ .

Answer 2

The maximum hight