Question

please answer the questions Q 4,5,7,9,10( I don't know if my answers are correct) please give explanation

40points

Answer 1

1) use formula,
t ={ u +√(u² +2gH)}/g

gt = u + √( u² +2gH)

9.8 × 6 = 19.6 +√(19.6²+2×9.8×H)

9.8 × 4 = √(19.6² +19.6×H)

take square both sides

9.8² × 16 = 19.6² +19.6 × H

9.8² ×16 -9.8×4 =19.6× H

9.8×3 = H

H = 58.8 m

2) use S = ut + 1/2at²
u = 0 and t =2 sec
S = 1/2a(2)² =2a

next two seconds
v = u + at =0 + 2a and t = 2 sec
S = ut + 1/2at²
=2a(2) +1/2a(2)² = 6a

next two seconds
v = u + at = 2a + 2a = 4a and t = 2 sec
S = ut + 1/2at²
= 8a + 2a = 10a

so, S1 : S2 : S3 = 2a : 6a : 10a = 1: 3 : 5


7) Let height of tower = H
and t is the time taken for complete journey of particle .

H = ut + 1/2gt²
H = 0 × t + 1/2gt²
H = 5t² -----(1)

let paticle strike after t second
then, distance , covered in tth second
55 = u + g(2t -1)/2
55 = 0 + 5(2t -1)
11 = 2t -1
t = 6

hence, height of tower = 5t² = 5×36 = 180 m

9) for ascending particle ,
v² = u² +2as
v² = u² + 2(-g)(-H)
v =√ ( u² +2gH)
Va = -√( u² +2gH)

for descending
Vb² = u²+ 2gH
Vb =√(v² + 2gH)

hence magnitude of Va = Vb


10) Let at H height fall the body then velocity becomes = 2× velocity of particle fall from h height .
use kinematics eqn
v² = u² +2as
V² = 0 + 2gh
V² = 2gh -------(2)

again ,
V'² = u² + 2as
= 0 + 2gH
(2V)² = 2gH
4V²= 2gH
4× 2gh = 2gH
H = 4h

hence , body when 3h above from its then velocity of body becomes zero

Answer 2

4) Let the height be "h"

Applying the formula,
t = u +√(u² +2gh)/g
gt = u + √( u² +2gh)
9.8 × 6 = 19.6 +√(19.6²+2×9.8×h)
9.8 × 4 = √(19.6² +19.6×h)
(9.8)² × 16 = 19.6² +19.6 × h (On squaring both the sides)

9.8² ×16 -9.8×4 =19.6×h
9.8×3 = h
h = 58.8 m

5) It is given that
u = 0
t = 2 seconds
We know that,
s = ut + 1/2at²
s = 1/2a(2)² =2a

For next 2 seconds,
v = u + at =0 + 2a
 t = 2 sec
s = ut + 1/2at²
=2a(2) +1/2a(2)² = 6a

For next 2 seconds,
v = u + at = 2a + 2a = 4a
 t = 2 sec
s = ut + 1/2at²
= 8a + 2a = 10a

Required ratio = 2a : 6a : 10a = 1 : 3 : 5


7) Let height of tower be "h"
Total time be "t"

h= ut + 1/2gt²
h = 0 × t + 1/2gt²
h = 5t² -----(i)

The particle strike after "t" second
So, the distance , covered in "t"th second --

55 = u + g(2t -1)/2
55 = 0 + 5(2t -1)
11 = 2t -1
t = 6

The required height of tower = 5t² = 5×36 = 180 m

9) Applying the formula,
v = u + at
v = 0 + (-a) t
v = -10 x 0.5
v= 5 m/s upwards

10)Let "h" be the height

v² = u² +2as
v² = 0 + 2gh
v² = 2gh -------(2)

again ,
v₁² = u² + 2as
= 0 + 2gh
(2v)² = 2gh
4v²= 2gh
4× 2gh = 2gh
H = 4h

Hence required distance is 3h.

The correct options to your query are :

4) D
5) C
7) B
9) B
10) A

Hope This Helps You!