A ball is dropped from a balloon going up at a speed of 7m/s.if the balloon was at the height 60m at the time of dropping the ball ,how long will the ball take in reaching the ground
Answers
Solution :
Initially ball is going upwards
Initial velocity=u= -7m/s
S=60m
a=g=10m/s2
From Second equation of motion
s=ut+1/2at²
60= - 7t +(1/2 )10 t²
60=-7t +5t²
5t² -7t -60=0
t=-7±√49-4x5(-60)/2x5
t=-7±√49+1200/10
t=-7±35.3/10
t=7+35.3/10 =4.2 sec [ as t cannot be negative]
∴ The ball will take 4.2 sec to reach the ground.
Answer:
4.24 s
Displacement will be
60
m
Explanation:
When the ball will be dropped,it will have an initial velocity of
7
m
s
upwards,
Now,let it will take time
t
to reach the ground,
so using
s
=
u
t
+
1
2
g
t
2
Here,
s
=
60
m
(as after going up,it will come down by that distance and will go down by the total displacement of
60
m
) and
,
u
=
−
7
m
s
(taking upward direction to be positive)
So,our equation becomes,
60
=
−
7
t
+
1
2
⋅
10
t
2
Solving,we get,
t
=
4.24
s
ALTERNATIVELY
So,let's calculate the time taken by it to reach the maximum height above the point of release.
using,
v
=
u
−
g
t
At its highest point,
v
=
0
So,
t
=
7
10
s
or
,
0.7
s
So,after this it will start falling downwards due to the effect of gravity,so net displacement will be
60
m
,as it will come down by that distance through which it went upwards after releasing it.
Now to come down by that distance it will take the same amount of time,so to reach its point of release from its highest point,it will take a total of
2
⋅
0.7
=
1.4
s
Let,it will take further time
t
to reach the ground after returning to its point of release,
So,using
s
=
u
t
+
1
2
g
t
2
Here,
s
=
60
m
,
u
=
7
m
s
(when returning to the point of release,it will gain the same velocity but direction will be reversed.)
So,our equation becomes,
60
=
7
t
+
1
2
10
t
2
Solving,we get,
t
=
2.834
s
So,total time required to come back to the ground is
1.4
+
2.834
=
4.24
s