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?
Home
/
From the foot of a tower 90m high, a stone is thrown up so as to reach the top of the tower. Two seconds
Verified answer
Explanation:
a=g=9.8m
let [tex]s_1[/tex]be distance from top of tower where two stone meet,so we have to find [tex]s_1[/tex]
part one➡
from equation of motion
[tex]s_1=ut-\frac{1}{2}[/tex]*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)
[tex]s_1=42t-\frac{1}{2}[/tex]*9.8t²
[tex]s_1[/tex]=42t-5t².............(1)
part two➡
here time of second stone is two second lass then first stone,mean (t-2)
[tex]s_2=ut+\frac{1}{2}[/tex]*9.8(t-2)²
[tex]s_2[/tex]=0t+5(t-2)²............(2)
since
[tex]s_1+s_2=90m[/tex]
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)
[tex]s_2[/tex]=5(t-2)²
[tex]s_2[/tex]=5(5-2)²
[tex]s_2[/tex]=5(3)²
[tex]s_2[/tex]=5*9
[tex]s_2[/tex]=45m
both stone will meet after 5second at 45m from top of tower
❤MARK IT AS BRILLIANT❤