Question

A brick is thrown upward from the top of a building at an angle of 25° to the horizontal and with an initial speed of 15 m/s. If the brick is in flight for 3.3 s, how tall is the building?

Answer 1

Concepts and reason

The concept required to solve this problem is projectile motion. To find the height of the building, calculate the displacement of the projectile in the y-direction after $3.3{\rm{ s}}$

from the time it was launched. For this initially calculate the y-component of the initial velocity. Finally find out the displacement of the projectile in y-direction, this is the height of the building.

Fundamentals

A projectile is anybody that is thrown in motion through space by action of force. During the flight of a projectile, the only force that is acting on the object is the force of gravity. This indicates that the horizontal velocity of the object is constant, and the vertical component of velocity is accelerating by the acceleration due to gravity constant.

Assume that the upward direction to be along the positive y-axis and the origin (0, 0) to coincide with the point of launch. Kinematic equation for vertical displacement is,

$\Delta y = {v_{0y}}t + \frac{1}{2}{a_y}{t^2}$ ,

Where ${v_{0y}}$ is the y-component of the initial velocity, ${a_y}$ is the acceleration due to gravity and $t$ is the time of flight.

$\begin{array}{l}\\{v_{0y}} = {v_0}\sin \theta \\\\{a_y} = - 9.8{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Calculate the y-component of initial velocity using equation,

${v_{0y}} = {v_0}\sin \theta$ .

Substitute 15 m/s for ${v_0}$ and ${\rm{25}}^\circ$ for $\theta$ .

$\begin{array}{c}\\{v_{0y}} = \left( {15\;{\rm{m/s}}} \right)\sin 25^\circ \\\\ = 6.345{\rm{ m/s}}\\\end{array}$

Calculate the height of the building, $H$ using the formula of vertical displacement of the projectile:

$H = \Delta y = {v_{0y}}t + \frac{1}{2}{a_y}{t^2}$

Substitute 6.345 m/s for ${v_{0y}}$ , $- 9.8{\rm{ m/}}{{\rm{s}}^2}$ for ${a_y}$ , and 3.3 s for t.

$\begin{array}{c}\\\Delta y = \left( {6.345{\rm{ m/s}}} \right)\left( {{\rm{3}}{\rm{.3 s}}} \right) + \frac{1}{2}\left( { - 9.8{\rm{m/}}{{\rm{s}}^2}} \right){\left( {3.3{\rm{ s}}} \right)^2}\\\\{\rm{ = - 32}}{\rm{.422 m}}\\\end{array}$

So, height of the building, $H = 32.42{\rm{ m}}$

Ans:

Height of the building, $H$ is $32.42{\rm{ m}}$ .