Question

187. A particle is projected with velocity 20 m/s, so that it just
clears two walls of equal height 10 m, which are at a
distance 20 m from each other. The time of passing between
the walls is: (Take g =10m/s2)
(a) 25
(b)-S
() 10/2s
(d) 2/10 s​

Answer 1

Answer:

ANSWER

Let t be the time of passing between the walls

S = vtS=vt

20 = 20 cos \theta t20=20cosθt -------------(i)

vertically

v^2 = u^2 +2aSv

2

=u

2

+2aS

v^2 = u^2 sin^2\theta -2g \times 10v

2

=u

2

sin

2

θ−2g×10

v^2 = u^2 sin^2\theta -20gv

2

=u

2

sin

2

θ−20g

v = u+atv =u+at

0 = u -gt0=u−gt

t = \dfrac{v}{g}t=

g

v

------------------(ii)

equating eq i and ii

\dfrac{1}{cos \theta} = 2\dfrac{v}{g}

cosθ

1

=2

g

v

squaring both sides and substituting v

\dfrac{g^2}{cos^2\theta} = 4u^2sin^2\theta - 80g

cos

2

θ

g

2

=4u

2

sin

2

θ−80g

let sin^2\thetasin

2

θ be x

g^2 = 16g^2x(1-x) -8g^2(1-x)g

2

=16g

2

x(1−x)−8g

2

(1−x)

9 =16x - 16x^2 +8x9=16x−16x

2

+8x

16x^2 -24x +9 =016x

2

−24x+9=0

(4x-3)^2 = 0(4x−3)

2

=0

x = \dfrac{3}{4}x=

4

3

sin^2\theta = \dfrac{3}{4}sin

2

θ=

4

3

\theta = \dfrac{\pi}{3}θ=

3

π

therefore time of passing between the walls

t = \dfrac{1}{cos\theta} = 2 st=

cosθ

1

=2s

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