A stone is thrown up from the ground with a certain speed . It crosses a certain point at times t1 and t2 from starting instant . Find the speed of projection and height of that point from ground ?
Answers
Let u & H be initial velocity and maximum height respectively so, u=
2gH
Now, for both time t
1
& t
2
displacement is h
h=ut−
2
1
gt
2
t
2
+
g
2u
t+
g
2h
=0
which is having two roots t
1
and t
2
t
1
+t
2
=
g
2u
ort
1
+t
2
=
g
2
2gH
⇒H=
8
g(t
1
+t
2
)
2
Let u & H be initial velocity and maximum height respectively so, u=
2gH
Now, for both time t
1
& t
2
displacement is h
h=ut−
2
1
gt
2
t
2
+
g
2u
t+
g
2h
=0
which is having two roots t
1
and t
2
t
1
+t
2
=
g
2u
ort
1
+t
2
=
g
2
2gH
⇒H=
8
g(t
1
+t
2
)
2
Let u & H be initial velocity and maximum height respectively so, u=
2gH
Now, for both time t
1
& t
2
displacement is h
h=ut−
2
1
gt
2
t
2
+
g
2u
t+
g
2h
=0
which is having two roots t
1
and t
2
t
1
+t
2
=
g
2u
ort
1
+t
2
=
g
2
2gH
⇒H=
8
g(t
1
+t
2
)
2
Let u & H be initial velocity and maximum height respectively so, u=
2gH
Now, for both time t
1
& t
2
displacement is h
h=ut−
2
1
gt
2
t
2
+
g
2u
t+
g
2h
=0
which is having two roots t
1
and t
2
t
1
+t
2
=
g
2u
ort
1
+t
2
=
g
2
2gH
⇒H=
8
g(t
1
+t
2
)
2
Let u & H be initial velocity and maximum height respectively so, u=
2gH
Now, for both time t
1
& t
2
displacement is h
h=ut−
2
1
gt
2
t
2
+
g
2u
t+
g
2h
=0
which is having two roots t
1
and t
2
t
1
+t
2
=
g
2u
ort
1
+t
2
=
g
2
2gH
⇒H=
8
g(t
1
+t
2
)
2
Let u & H be initial velocity and maximum height respectively so, u=
2gH
Now, for both time t
1
& t
2
displacement is h
h=ut−
2
1
gt
2
t
2
+
g
2u
t+
g
2h
=0
which is having two roots t
1
and t
2
t
1
+t
2
=
g
2u
ort
1
+t
2
=
g
2
2gH
⇒H=
8
g(t
1
+t
2
)
2
vvLet u & H be initial velocity and maximum height respectively so, u=
2gH
Now, for both time t
1
& t
2
displacement is h
h=ut−
2
1
gt
2
t
2
+
g
2u
t+
g
2h
=0
which is having two roots t
1
and t
2
t
1
+t
2
=
g
2u
ort
1
+t
2
=
g
2
2gH
⇒H=
8
g(t
1
+t
2
)
2
Aa
Hope it helps you ,