Question

From the foot of a tower 90m high, a stone is thrown up so as to reach the top of the tower. Two seconds later another stone is dropped from the top of the tower. When will the stone meet?

Answer 1

Explanation:

• here, s=90m
• since acceleration produce in stone in both cases is equal to acceleration due to gravity.

a=g=9.8m

let $s_1$be distance from top of tower where two stone meet,so we have to find $s_1$

part one➡

from equation of motion

$s_1=ut-\frac{1}{2}$*at²..........(a)

(here g is negative)

here we have to find initial velocity

s=90m

since when stone reach top its final velocity become zero, so v=0

from equation

v²-u²=2as

0²-u²=2(-9.8)90

u²=1764

u=42m/s

put in equation (a)

$s_1=42t-\frac{1}{2}$*9.8t²

$s_1$=42t-5t².............(1)

part two➡

here time of second stone is two second lass then first stone,mean (t-2)

$s_2=ut+\frac{1}{2}$*9.8(t-2)²

$s_2$=0t+5(t-2)²............(2)

since

$s_1+s_2=90m$

42-5t²+5(t-2)²=90m

42t-5t²+5(t²+4-4t)=90

42t-5t²+5t²+20-20t=90

22t+20=90

22t=110

t=5sec

distance where they meet

from equation (2)

$s_2$=5(t-2)²

$s_2$=5(5-2)²

$s_2$=5(3)²

$s_2$=5*9

$s_2$=45m

both stone will meet after 5second at 45m from top of tower

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