Question

A ball thrown vertically up from the top of a
ort tower reaches the ground in 12 s. Another ball
led thrown vertically downwards from the same posi-
tion with the same speed takes 4 s to reach the
ground. Find the height of the tower. (Take g =
che
10 m 2
s,
1​

Answer 1

For both the bodies,

The displacement and the final velocity are equal.

Also, their initial velocities are equal but opposite in direction.

From the formulas  $s=ut+\frac{1gt^{2} }{2} \ and\ u=v-gt$.

h= $vt-\frac{1gt^{2} }{2}$

This can be represented by the quadratic equation  $\frac{1gt^{2} }{2}-vt+h=0$.

This quadratic equation is t.

Given, $t_{1} =12s\ t_{2} =4s$

If $\alpha$ and $\beta$ are the two roots of the equation.

Then $\alpha +\beta =\frac{-b}{a} \\\alpha \beta =\frac{c}{a}$

Sum of roots=$t_{1} +t_{2} =\frac{v}{g/2}$

velocity(v)=$16\times 5=80m/s$

Products of roots=$t_{1}t_{2}=\frac{h}{g/2}$

Height(h)= $12\times 4\times 5=240m$

Hence, the  the height of the tower is 240m.