Question

Q body is thrown vertically upwards at t=0 . It is at a height 80m at two instants t1 and t2 , then t1×t2 is ?​

Answer 1

Answer : 16

Hope the attached image helps....

Answer 2

Answer:

The product of the two instants of time is 16.

Explanation:

Given the height reached by an object, $h=80m$

A body is thrown vertically upwards, therefore $g=-10m/s^{2}$

From the equation of motion, $s=ut+\frac{1}{2} at^{2}$

we get $h=ut-\frac{1}{2} gt^{2}$

$\implies 80=ut-\frac{1}{2} \times 10t^{2}$

$\implies 80=ut-5t^{2}$

$\implies5t^{2}-ut+ 80=0$

Comparing this with the standard form of quadratic equation, $ax^{2}+bx+ c=0$ we have here, $a=5, b=-1, c=80$.

The product of the roots of a quadratic equation is given by $\frac{c}{a}$ .

Let the two roots of the equation be $t_{1}~\&~ t_{2}$.

The product of the roots, $t_{1}~\times~ t_{2}=\frac{80}{5} =16$.

Therefore, the product of the two instants of time is 16.