A Tennis ball is released from a height h and after freely falling on a wooden floor it rebounds and reaches height h2 . The velocity versus height of the ball during its motion may be represented graphically by:

(graphs are drawn schematically and on not to scale)

Option 1 -

(a)

Option 2 -

(b)

Option 3 -

(c)

Option 4 -

(d)

0 7 Views | Posted 2 months ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    2 months ago
    Correct Option - 3


    Detailed Solution:

    V 2 = U 2 + 2 g S V 2 = 0 + 2 g ( h - y ) V 2 = 2 g h - 2 g y V = 2 g h - 2 g y

Similar Questions for you

A
alok kumar singh

Let 't' be the time taken by B to meat A.                                                                

for A, u = 0

t = [2 + t]

h = 80m

a = 10 m/sec2

h = ut +    1 2 a t 2

80 = 0 * t + 1 2 ( 1 0 ) ( 2 + t ) 2

16 = (2 + t)2

t + 2 = 4

 t = 2sec

Now for B

h = ut +   1 2 a t 2

           

...more
P
Payal Gupta

Let ‘t’ be the time taken by B to meat

A.

for A, u = 0

t = [2 + t]

h = 80m

a = 10 m/sec2

h = ut + 12at2

80 = 0 × t + 12 (10) (2+t)2

16 = (2 + t)2

t + 2 = 4

t = 2sec

Now for B

h = ut + 12at2

80=u×2+12×10t2

80 = 2u + 5 × 2 × 2

80 = 2u + 20

2u = 60

u = 30m/sec

P
Payal Gupta

tanα=vmaxt1=a1

tanβ=vmaxt2=a2

a1a2=t2t1t1t2=a2a1

V
Vishal Baghel

h = 39.2 + 19.6 = 58.8 m

Height above tower

h'=u22g=19.6×19.62×9.8

= 19.6

As, k5=78.4k=392

A
alok kumar singh

K=QrΔxΔAT

ML2T-2 (L)L2 (θ) (T)M1L1-T-3θ-1

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