Prove that if a body is thrown vertically upwards with V then the time ofa is even to the time of decent
Answers
Acceleration
Distance
Final Velocity
v = u + a*t
Initial Velocity u
m/s
Final Velocity v
m/s
Time t
s
Acceleration a
m/s²
RESET VALUES
ANSWER
Let,
u = Initial velocity
v = Final velocity
h = Height attained
a = acceleration
t = time
Ascent:
Ball thrown upward with velocity u So,v=0 at maximum height h. Also a=−g
Now, using v=u+at
0=u−gt
a
t
a
=
g
u
.....1
Where t
a
= time of ascent
Also , by using v
2
=u
2
+2as
0=u
2
−2gh
u=
2gh
........2
From equation 1 and 2 , we get ,
t
a
=
g
2h
.......A
Descent :
Ball comes down from height h , So , u=0 and a=g .
So , using v=u+at
v=gt
d
.........3
Where t
d
= Time of descent
Now , using v
2
=u
2
+2as
v
2
=0+2gh
v=
2gh
..........4
From equations 3 and 4 , we get ,
t
d
=
g
2h
.....B
From equations A and B , we see that t
a
=t
d