Question

A ball thrown vertically upwards with a speed of 19.6 m/s from the top of a tower returns to the earth in 6s. What is the height of the tower?

Answer 1

We know that in the case of a vertically projected body, velocity is maximum at the projection point, velocity goes on decreasing and becomes minimum (zero) at the highest point.

Formula used:
Height of the tower, $H = \dfrac{1}{2}g{t^2} - ut$
$u$ is the initial velocity of the body,
Time of travel=$t$

Complete step by step solution:
Consider a tower of height H. suppose a body is projected upwards vertically with initial velocity u from the top of the tower. Suppose it reaches a displacement x above the tower and thereafter reaches the foot of the tower.

Let t be the total time of travel.
Now considering the total path of the body, the motion parameters are as follows:
Initial velocity of the body=u
Net displacement of the body $= S = + x - x - H = - H$
Time of travel=t
We know that, from kinematic equation of motion,
$S = ut + \dfrac{1}{2}a{t^2}$
Here,
$a = - g,S = - H,u = u,t = t$
Therefore, we get
$- H = ut - \dfrac{1}{2}g{t^2}$
$\therefore H = \dfrac{1}{2}g{t^2} - ut$ ………………………… (1)
From this above equation we can calculate the height of the tower:
Given, initial velocity=19.6m/s
Time=6s
$g = 9.8m{s^{ - 2}}$
Now substitute these values in equation (1),
$H = \dfrac{1}{2} \times 9.8 \times {6^2} - 19.6 \times 6$
$H = 58.8m \approx 59m.$
Therefore, the height of the tower is 59m.
State and prove work-energy theorem:
It states that the change in kinetic energy of a body is equal to the work done on it by the net force.
Consider a body of mass m moving with velocity ${{\text{v}}_{\text{1}}}$. The initial kinetic energy of the body is given by, $\dfrac{{\text{1}}}{{\text{2}}}{\text{m}}{{\text{v}}_{\text{1}}}^{\text{2}}$
Let the body be subjected to a constant force F so that its velocity increases to ${{\text{v}}_{\text{2}}}$. Now the final kinetic energy of the body is given by, $\dfrac{{\text{1}}}{{\text{2}}}{\text{m}}{{\text{v}}_{\text{2}}}^{\text{2}}$
From Newton’s second law of motion,
$F = ma$
Acceleration of the body is, $a = \dfrac{F}{m}$ ………………………. (2)
Using the kinematic equation of motion, ${v^2} = {u^2} + 2as$ , we get
Here initial velocity is ${{\text{v}}_{\text{1}}}$ and final velocity is ${{\text{v}}_{\text{2}}}$then,
${v_2}^2 = {v_1}^2 + 2as$
${v_2}^2 = {v_1}^2 + 2\left( {\dfrac{F}{m}} \right)s$ from (2)
Multiply above equation by$\dfrac{1}{2}m$ , we get,
$\dfrac{1}{2}m{v_2}^2 = \dfrac{1}{2}m{v_1}^2 + \dfrac{1}{2}m2\left( {\dfrac{F}{m}} \right)s$
Simply above equation,
$\dfrac{1}{2}m{v_2}^2 = \dfrac{1}{2}m{v_1}^2 + Fs$
Here the product of force and displacement is given by work done. That is, $W = Fs$
Now we get, $\dfrac{1}{2}m{v_2}^2 = \dfrac{1}{2}m{v_1}^2 + W$
$\dfrac{1}{2}m{v_2}^2 - \dfrac{1}{2}m{v_1}^2 = W$
$\therefore W = {K_f} - {K_i} =$Change in kinetic energy
Therefore, $W = \Delta K$

Hence work-energy theorem is proved.

Note:
If a particle is projected vertically upward with the velocity u then acceleration $a=-g$.
If a particle is thrown vertically downward with the velocity u then acceleration $a=+g$.
Work done is a product of force and displacement.