Physics

Session

Kinematics - 3

Session Objectives

Problems ?

Free fall under gravity

Free Fall Under Gravity

- g

g is constant = - g

-y

Motion under gravity is along y-axis. Downward is negative

One dimensional equations of motion under gravity

v = u - gt

v² = u² - 2gs

21s ut gt

2

n

2n 1S u g

2

Distance traveled in nth second of fall

Class Exercise

Class Exercise -1Which of the following graphs defines the motion of an object thrown vertically up with a velocity V from a height H above the ground?

Tim e

Dis

plac

emen

t

t 2t

(a)

Tim e

Dis

plac

emen

t

t 2t

(b)

Tim e

Dis

plac

emen

t

t1 t2

t2 > 2t1

(c)

Time

Dis

plac

emen

t

t1 t2

t2 > 2t

(d)

Solution

Tim e

Dis

plac

emen

t

t1 t2

t2 > 2t1

Hence answer is (c)

Class Exercise -2

A stone is dropped from a rising balloon at a height of 76 m above the ground and reaches the ground in 6 s. What is the approximate velocity of the balloon when the stone was dropped?

52(a) m / s upwards

3

52(b) m / s downwards

3

(c) 3 m/s (d) 9.8 m/s

Solution

u Initia l position

Final position

76 m

If u is the velocity of balloon in the upward direction when the stone was dropped, then

2176 u(6) 10 (6 )

2

–76 = 6u – 180

52u m/s

3

Class Exercise - 3A particle moves in a straight line with a velocity v = –3t2 + 2t + 5 ms–1, where t is the time. Assuming x(t) = 0 at t = 0, find

(i) acceleration a(t), at t = 2 s,

(ii) position of particle at t = 5 s.

Solution

a(2) = –6 × 2 + 2 = –10 ms–2

(ii)dx

vdt

0

t

t

x v t dt

5

2

0

x 5 3t 2t 5 dt

53 2

0t t 5t = –75 m

dva 6t 2

dt (i)

Class Exercise - 4

What is the time taken by an object thrown vertically upwards at 50 m/s to attain a displacement which is half of the maximum displacement?

(Take g = 10 m/s2)

Solution

2

maxu 50 50

S 125 m2g 2 10

max

2

s 62.5 m

2162.5 50t – 10 t

2

212.5 10t – t

2t2 – 20t + 25 = 0

40 400 – 200t

4

40 200 20 5 2 20 – 5 2 s or s

4 2 2

Class Exercise - 5

For an object moving in one dimension, find a if v = 3x2 + 4x m/s.

2v 3x 4x

dv dv dx

dx dt dt

(6x 4) a v

2

26x 4

a m/s3x 4

Solution :

Class Exercise - 6

a a

u u

v = ?m

A mass m is being pulled up using two strings passing over two pulleys so that the downward velocities of the strings are equal to u. (see figure). What is the speed v of the mass ?

Solution

v cos = u

uv

cos

a a

v

u uu u

m

Class Exercise - 7

Two objects are thrown, one upwards and one downwards with a speed u from the top of a tower. If the time of flights are t1 and t2, find the time of flight of an object dropped from the tower.

SolutionThe time of flight to start from P and come back to that point is t1 – t2 for the ball thrown upwards.

u

u

P

t – t1 2

1 22u

(t – t )g

1 2g

u (t – t )2

22 2

1Height of the tower H ut gt

2

2

1 2 2 22

gt t gt 1– gt

2 2 2 1 2gt t

2

Hence, time of flight for a ball that is dropped

2H

g 1 2t t s

Class Exercise - 8An object is thrown upwards from the top of a tower with a velocity v1. At the same instance another object is thrown from the bottom of the tower with a velocity v2. If they meet at the top of the tower, find the ratio v1 : v2. (Assume v2 is just enough to reach the top of the tower.)

Solution

11

2vt

g

22

vt

g

v1 : v2 = 1 : 2

Given that t2 = t1

v2

v1 t1

t2

Class Exercise - 9

v 10 i 6 j m/s ; u 2 i m/s. Find a if t 3s.

Along X-axis: x10 – 2 8

a3 3

Along Y-axis: y6 – 0

a 23

x ya a i a j

28

i 2 j m/s3

Solution :

Class Exercise - 10

2v 10 i 6 j m/s ; u 2 i m/s ; a x i +y j m/s

Find x and y if t = 5 s.

Use the same method as question 9.

8 6

x m; y m5 5

Solution :

Thank you