Question

An object is thrown upwards with an initial velocity of 17 m/sec from a building with 12 m height. It is at a height of $ S = 12 + 17t - 5{t^2} $ from the ground after a flight of 't' seconds. Find the time taken by the object to touch the ground.

Answer 1

Hint: Velocity with which the object travels is 17 m/sec and the height of the building is 12m. After a time t seconds, we have the equation of height as $ S = 12 + 17t - 5{t^2} $ in terms of velocity (17), height (12) and acceleration (5). As the height is relative to the ground, when an object touches the ground the height will be 0 m. So the value of S in the equation $ S = 12 + 17t - 5{t^2} $ will be zero when the object touches the ground. Find the value of t using the below formula of quadratic equation.
Formula used:
When a quadratic equation is in the form of $ a{x^2} + bx + c = 0 $ where a is not equal to zero, the value of x will be $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ where a, b and c are the coefficients. This formula is called a quadratic formula.

Complete step-by-step answer:
We are given that an object is thrown upwards with an initial velocity of 17 m/sec from a building with 12 m height and it is at a height of $ S = 12 + 17t - 5{t^2} $ from the ground after a flight of 't' seconds.
When the object touches the ground, the height S will be zero.
Therefore, $ S = 0 $
But we already have that $ S = 12 + 17t - 5{t^2} $
This gives us $ 0 = 12 + 17t - 5{t^2} $
 $ \Rightarrow 12 + 17t - 5{t^2} = 0 $
As we can see the above equation is a quadratic equation and when we compare the above equation with $ a{x^2} + bx + c = 0 $ , we get $ a = - 5,b = 17,c = 12 $
The value of $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ , which means the value of $ t = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
Substitute the values of a, b and c to find the value of t
 $ \Rightarrow t = \dfrac{{ - \left( {17} \right) \pm \sqrt {{{17}^2} - 4\left( { - 5} \right)\left( {12} \right)} }}{{2\left( { - 5} \right)}} $
 $ \Rightarrow t = \dfrac{{ - 17 \pm \sqrt {289 + 240} }}{{ - 10}} = \dfrac{{17 \pm \sqrt {529} }}{{10}} = \dfrac{{17 \pm 23}}{{10}} $
 $ \Rightarrow t = \dfrac{{17 + 23}}{{10}},t = \dfrac{{17 - 23}}{{10}} $
 $ \Rightarrow t = \dfrac{{40}}{{10}},t = \dfrac{{ - 6}}{{10}} $
 $ \Rightarrow t = 4sec,t = - 0.6sec $
We have got 2 values for t, but one is positive and one is negative.
Time cannot be negative.
Therefore the time taken by the object to touch the ground is 4 seconds.
So, the correct answer is “4 seconds”.

Note: Quadratic equations can also be factored instead of using the above formula to find the values of x. When the equations cannot be factored we can use quadratic formulas. Using quadratic formulas, we may get real values and imaginary values. The no. of solutions of an equation depends upon the highest degree of the variable. If the highest degree is 2, it will have 2 solutions; if the highest degree is 3 it will have 3 solutions and so on.