1 Crore+ students have signed up on EduRev. Have you? 
2. Motion under gravity :
I format : (When a body is thrown vertically upward)
It includes two types of motion
(i) Deaccelerated motion from A to B because the direction of velocity and acceleration is opposite. So speed
decreases
(ii) Accelerated motion from B to C because the direction of velocity and acceleration is same (downward). So speed increases
(a) Time of flight :
It is the time taken by the particle to reach the ground. If the particle is thrown vertically upward with initial velocity u then
u_{i }= u
a = – g (take downward direction negative)
from equation
S = ut + ⇒ S_{net }= 0 (when particle again reaches the ground)
t = T (time of flight)
(b) Maximum Height :
from v^{2} = u^{2} + 2as
at maximum height v = 0, s = H_{max}
⇒ 0 = u^{2} –2 gH_{max } ⇒
(c) Final velocity
from v = u + at
v = v_{f} a = – g
v_{f }= – u
i.e. the body reaches the ground with the same speed with which it was thrown vertically upwards as it thrown vertically upward.
(d) Time to reach half of the maximum height :
u_{i} = u a = – g
from
H = 2ut – gt^{2} ⇒ gt^{2 }– 2ut + H = 0
⇒
(.....1)
Equation 1 gives two value of time which corresponds to
t_{1} = (from ground to H_{max}/2 in upward motion)
t_{2} = (from ground to H_{max}/2 in downward motion)
(e) Time to reach any general height h
Let us assume that particle reaches from A to B in time t_{1 }& from A to C is time t_{2}.
gt^{2} – 2ut + 2h = 0
⇒ t_{1} + t_{2} = T (Time of flight)
II Format (Free fall) :
A body released near the surface of the earth is accelerated downward under the influence of force of gravity.
(a) Time of Flight :
S = – H, u = 0, a = – g
t = T (Let assume)
⇒ – H = (0)T –
(b) Final Velocity when body reaches the ground
from v^{2} – u^{2} = 2as
s = – H v = v_{f } u = 0 a = – g
⇒ – 0 = 2 (–g) (–H) ⇒ v_{f} =
Ex.17 A ball is thrown vertically upwards with a velocity u from the ground. The ball attains a maximum height H_{max}. Then find out the time and displacement at which ball have half of the maximum speed.
Sol. Maximum speed of the ball is u At point B and C ball have speed u/2 but direction
is opposite so from
v = u + at
Let t_{1} is the time taken by the ball from point A to B and t_{2} is the time taken by the ball from A to C
Ex.18 A ball thrown vertically upwards with a speed of 19.6 ms^{–1} from the top of a tower returns to the earth in 6 s. Find the height of the tower.
Sol. Here u = 19.6 ms^{–1}
g = –9.8 ms^{–2}
Net displacement, s = – h
Negative sign is taken because displacement is in the opposite direction of initial velocity.
∴ – h = 19.6 × 6 + 1/2 × (–9.8) × 62
= 117.6 – 176.4 = –58.8
or h = 58.8 m
Ex.19 A ball is thrown vertically upwards with a velocity of 20 ms^{–1} from the top of a multistoreyed building. The height of the point from where the ball is thrown is 25 m from the ground. (i) How high will the ball rise and (ii) how long will it be before the ball hits the ground?
Sol. (i) Here u = +20 ms^{–1}, g = –10 ms^{–2 }
At the highest point, v = 0
Suppose the ball rises to the height h from the point of projection.
As v^{2} – u^{2} = 2gs
∴ 02 – 202 = 2 × (–10) × h or h = + 20 m.
(ii) Net displacement, s = –25 m
Negative sign is taken because displacement is in the opposite direction of initial velocity.
As s = ut +
or 5t^{2} – 20t – 25 = 0 or t^{2} – 4t – 5 = 0
or (t+ 1) (t – 5) = 0
As t ≠ –1, so t = 5s.
Ex.20 A ball thrown up is caught by the thrower after 4s. How high did it go and with what velocity was it thrown ? How far was it below the highest point 3 s after it was thrown?
Sol. As time of ascent = time of descent
∴ Time taken by the ball to reach the highest point = 2 s
For upward motion of the ball : u = ?, v = 0, t = 2s, g = – 9.8 ms–2
As v = u + gt
∴ 0 = u – 9.8 × 2
or u = 19.6 ms^{–1}
Maximum height attained by the ball is given by
Displacement of the ball in 3 s,
s = 19.6 × 3 + × (–9.8) × 32 = 58.8 – 44.1 = 14.7 m
Distance of the ball from the highest point 3 s after it was thrown
= 19.6 – 14.7 = 4.9 m.
Ex.21 A balloon is ascending at the rate of 9.8 ms–1 at a height of 39.2 m above the ground when a food packet is dropped from the balloon. After how much time and with what velocity does it reach the ground?
Take g = 9.8 ms^{–2}.
Sol. Initially the food packet attains the upward velocity of the balloon, so
u = 9.8 ms^{–1}, g = 9.8 ms^{–2} , s = –39.2 m
Here s is taken negative because it is in the opposite direction of initial velocity.
Using, s = ut + 1/2 gt^{2} , we get
– 39.2 = 9.8 t – 1/2× 9.8 t^{2}
or 4.9 t^{2} – 9.8 t – 39.2 = 0 or t^{2} – 2t – 8 =0
or (t – 4) (t + 2) = 0 or t = 4s or – 2s
As time is never negative, so t = 4s.
Velocity with which the food packet reaches the ground is
v = u + gt = 9.8 – 9.8 × 4 = – 29.4 ms^{–1}.
Negative sign shows that the velocity is directed vertically downwards.
When a particle is dropped then it will automatically attains the velocity of the frame at that time.
Ex.22 Two balls are thrown simultaneously, A vertically upwards with a speed of 20 ms^{–1} from the ground, and B vertically downwards from a height of 40 m with the same speed and along the same line of motion. At what points do the two balls collide? Take g = 9.8 ms^{–2}.
Sol. Suppose the two balls meet at a height of x from the ground after time t s from the start.
For upward motion of balls A :
u = 20 ms^{–1}, g = – 9.8 ms^{–2}
s = ut + 1/2gt^{2}
x = 20 t – 1/2 × 9.8 t^{2} = 20t – 4.9 t^{2} ...(i)
For downward motion of ball B,
40 – x = 20 × t + 1/2 × 9.8 t^{2}
= 20t + 4.9 t^{2} ... (ii)
Adding (i) and (ii), 40 = 40 t or t = 1 s
From (i), x = 20 × 1 – 4.9 × (1)^{2} = 15.1 m
Hence the two balls will collide after 1 s at a height of 15.1 m from the ground.
258 videos633 docs256 tests

258 videos633 docs256 tests
