A particle is projected in upward direction with initial velocity v. The height attained by the particle in time is H =vt-gt^2/2.
where g is acceleration due to gravity. Mark the correct statement(s).
(a) H is maximum at t = 10
(b) H is zero at t =2v/g
(c) The maximum value of H is v^2/2g
(d) His maximum at t = 2v/g
Answers
Answer:
a. Consider the motion from A to B:
s = +H (final point lies above the initial point), initial velocity = u, final velocity v = 0.
Let the time taken to go from A to be t
1
Using v = u - gt, we get
0 = u - gt
1
⇒ t
1
=
g
u
Using s = ut - (1/2)gt
2
,
H = u(
g
u
) -
2
1
g (
g
u
)
2
=
2g
u
2
So the maximum height attained is H =
2g
u
2
Consider the return motion from B to A:
s = -H (final point lies below the initial point)
u = 0 (at point B, velocity is zero)
Let time taken to go from B to A be t
2
. We have
t
2
=
g
2H
=
g2g
2u
2
=
g
u
Here t
1
is known as the time of ascent and t
2
is known as the time of descent. We can see that
Time of ascent = Time of descent =
g
u
Total time of flight, T = t
1
+ t
2
=
g
u
Here time of flight is the time for which the particle remains in the air.
Alternativemethodtofindthetimeofflight:
a. Consider the motion from A to B:
s = 0 (initial and final points are same)
Initial velocity = u, time taken =T
Using s = ut - (1/2)gt
2
, we have 0 =uT - (1/2)gT
2
⇒ T =
g
2u
b. Magnitude of velocity on returning the ground Will be same as that of initial velocity but direction will be opposite.
Proof: Let v be the velocity on reaching the ground. Then from previous formulae, we get
v=−
2gH
=−
2g
2g
u
2
⇒ v = - u, hence proved.
c. Displacement = 0, distance travelled = 2H =
2g
u
2
