CBSE BOARD XII, asked by sunilv9717, 14 days ago

A particle is projected at a speed of 4 ms at an angle of 15° on a horizontal surface. What is the speed with which it hits the surface on landing? (in ms"') RESPONSE AREA?

Answers

Answered by hasinikasa2010
1

Answer:

Correct option is

C

15.4 m

∴v=v

x

i

^

+v

y

j

^

v=i×u

x

cosθ

i

^

+u

x

×cosθ×tan(θ/2)

j

^

v

2

=u

2

cos

2

θ+u

2

cos

2

θ×tan

2

θ/2

v

2

=u

2

cos

2

θ(1+tan

2

θ/2)

v

2

=u

2

cos

2

θ×sec

2

θ/2

∴ Radius of curvature =

gcosθ/2

u

2

cos

2

θ×sec

2

θ/2

=

4×3×10×

3

400×1×4×2

3

3

80

Answered by knameera861
1

Explanation:

solution: Correct option is

D

40

3

m/s

Time of flight =

g

2usinθ

=

10

2×80×sin30

o

=8

As particle during journey at t=2sec and at t=6sec will be at same height and hence vertical displacement is zero.

Horizontal displacement =(ucosθ)×(6−2)

=80×

2

3

×4

=160

3

m

∴ Average velocity =

4

160

3

=40

3

m/sec

solution

expand

Was this answer helpful?

upvote

56

downvote

2

MORE FROM CHAPTER

Probability

View chapter

>

QUESTION SETS

JEE Mains Questions

14 Qs

>

JEE Advanced Questions

14 Qs

>

Easy Questions

99 Qs

>

Medium Questions

647 Qs

>

Hard Questions

254 Qs

>

Enjoy a better experience

Similar questions