Question

A ball is dropped from the roof of a tower of height h. The total distance covered by it in the last second of its motion is equal to the distance covered by it in the first three seconds. What is the value of $\mathrm{h} ?\left(\mathrm{g}=10 \mathrm{m} / \mathrm{s}^{2}\right)$

Answer 1

Hint: We should know that velocity is defined as the rate change of displacement per unit time. Speed in a specific direction is also known as velocity. Velocity is equal to displacement divided by time. Speed, being a scalar quantity, is the rate at which an object covers distance. The average speed is the distance which is a scalar quantity per time ratio. On the other hand, velocity is a vector quantity; it is direction-aware. An object which moves in the negative direction has a negative velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion in this case. Based on this we have to solve this question.

Complete step by step answer
A ball is dropped from the roof of a tower height $=\mathrm{h}$
Total distance covered in the last seconds of its motion is to be found out by us.
The distance covered in first 3 second will be given by the formula:
$\mathrm{s}=\dfrac{1}{2} \mathrm{at}^{2}$
Now we have to put the values to get the expression as:
$=\dfrac{1}{2} \times 10 \times 3^{2}$
$=45$
If the ball takes in second to fall to ground. The distance covered in nth second will be given by the formula as:
$\mathrm{s}_{\mathrm{n}}=\mathrm{u}+\dfrac{\mathrm{g}}{2}(2 \mathrm{n}-1)$
Now we have to put the values to get the expression as:
$=0+\dfrac{10}{2}(2 \times \mathrm{n}-1)$
$=10 \mathrm{n}-5$
Therefore, we can write now, $45=10 \mathrm{n}-5$
$10 \mathrm{n}=50$
$\mathrm{n}=5$
Therefore, $h=\dfrac{1}{2}=\dfrac{1}{2}\text{g}{{\text{t}}^{2}}$
Now we have to put the values the in the expression to get as:
$=\dfrac{1}{2} \times 10 \times 25$
$=125 \mathrm{m}$

Hence the height will be 125 m.

Note: We should know that if an object's speed or velocity is increasing at a constant rate then we say it has uniform acceleration. The rate of acceleration is constant. If a car speeds up then slows down then speeds up it doesn't have uniform acceleration. The instantaneous acceleration, or simply acceleration, is defined as the limit of the average acceleration when the interval of time considered approaches 0. It is also defined in a similar manner as the derivative of velocity with respect to time. If an object begins acceleration from rest or a standstill, its initial time is 0. If we get a negative value for acceleration, it means the object is slowing down. The acceleration of an object is its change in velocity over an increment of time. This can mean a change in the object's speed or direction. Average acceleration is the change of velocity over a period of time. Constant or uniform acceleration is when the velocity changes the same amount in every equal time period.