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**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 multi-storeyed 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.

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