
A body falling from a high Minaret travels \[40m\] in the last \[\;2\;\] seconds of its fall to the ground. The height of Minaret in meters is
( take \[g = 10m/{s^2}\])
A.\[60\]
B.\[45\]
C.\[80\]
D.\[50\]
Answer
484.8k+ views
Hint: The key to answering this question is provided by the equation of motions. The physical behavior of the object includes position, velocity, acceleration, etc. These physical behaviors of an object can be explained using equations of motion. We need to divide the question into two parts: the last two seconds when the object falls in the ground and before two seconds. Substituting the equation of motion in both cases we can find the height of the minaret.
Complete answer:
The second equation of motion gives us the formula,
\[s = ut + \dfrac{1}{2}a{t^2}\]
Here \[s\] is the displacement of the object
\[u\] is the initial velocity.
\[a\] is the acceleration of the body.
\[t\] is the time taken.
When a body falls freely under the influence of gravity and when its initial velocity is zero and acceleration is only due to gravity and g remains constant when height \[\;h\] is not a large quantity, the equation of motion will be reduced to the form,
Let \[h\] be the total height and \[t\] be the total time taken.
Therefore for the second case is when Minaret travels \[40m\] that is \[h - 40\] in the last \[\;2\;\] seconds that is \[t - 2\] of its fall to the ground. Substituting in the above equation,
\[h - 40 = 5({t^2} + 4 - 4t)\]………. (1)
For total height and the total time is taken we can write the above equation as,
………. (2)
\[h = 5{t^2}\] …… (3)
Substituting (3) in (1)
\[5{t^2} - 40 = 5\left( {{t^2} + 4 - 4t} \right)\]
\[ \Rightarrow 5{t^2} - 40 = 5{t^2} + 20 - 20t\]
\[ \Rightarrow 20t = 60\]
\[ \Rightarrow t = 3s\] ……… (4)
Substituting (4) in (1)
\[h = 5{(3)^2}\]
\[h = 45m\]
Therefore the correct option is B.
Note:
Equations of motion are used to describe the physical system’s behavior using a set of mathematical functions. They are usually used to calculate the components of motion. It is used to calculate different parameters involved in a motion such as velocity, time, acceleration, etc.
Complete answer:
The second equation of motion gives us the formula,
\[s = ut + \dfrac{1}{2}a{t^2}\]
Here \[s\] is the displacement of the object
\[u\] is the initial velocity.
\[a\] is the acceleration of the body.
\[t\] is the time taken.
When a body falls freely under the influence of gravity and when its initial velocity is zero and acceleration is only due to gravity and g remains constant when height \[\;h\] is not a large quantity, the equation of motion will be reduced to the form,
Let \[h\] be the total height and \[t\] be the total time taken.
Therefore for the second case is when Minaret travels \[40m\] that is \[h - 40\] in the last \[\;2\;\] seconds that is \[t - 2\] of its fall to the ground. Substituting in the above equation,
\[h - 40 = 5({t^2} + 4 - 4t)\]………. (1)
For total height and the total time is taken we can write the above equation as,
………. (2)
\[h = 5{t^2}\] …… (3)
Substituting (3) in (1)
\[5{t^2} - 40 = 5\left( {{t^2} + 4 - 4t} \right)\]
\[ \Rightarrow 5{t^2} - 40 = 5{t^2} + 20 - 20t\]
\[ \Rightarrow 20t = 60\]
\[ \Rightarrow t = 3s\] ……… (4)
Substituting (4) in (1)
\[h = 5{(3)^2}\]
\[h = 45m\]
Therefore the correct option is B.
Note:
Equations of motion are used to describe the physical system’s behavior using a set of mathematical functions. They are usually used to calculate the components of motion. It is used to calculate different parameters involved in a motion such as velocity, time, acceleration, etc.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

