Question

The benches of a gallery in a cricket stadium are 1 m wide and 1 m high. A batsman strikes the ball at a level one m above the ground and hits a mammoth sixer. The ball starts at 35 m/s at an angle of 53° with the horizontal. The benches are perpendicular to the plane of motion and the first bench is 110 m from the batsman. On which bench will the ball hit ? Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics

Answer 1

Solution:
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Distance of first bench =110m
Let us assume that the ball strikes the nth bench height above batsmen strike level=n-1
Horizontal distance=( 110+ n) m
h=ut-1/2 gt²
n-1=35sin 53°t-1/2 x 9.8 t²
4.9t²-28t+(n-1)=0
t=28 ±√28²-4(4.9)(n-1)/2x4.9
t=28/9.8 + √784-196(n-1)/9.8 [ we will only positive value]
t=2.86 +√8.16-0.204(n-1) 
Horizontal distance travelled =35cos53°t=110+n
21.06[2,86+√8.16 -0.204(n-1)]=110+n
60.24+21.06√8.16 -0.204(n-1)]=110+n
21.06√8.16 -0.204(n-1)]=49.76+n
squaring on both sides ,
444(8.16-0.204(n-1)=24.76+99.5214 +n²
n²+99.52n+90.58n+2476-3623+90.58=0
n²+190.1n-1056.42=0
Solving for n we get, n=5.4

That means the ball will cross 5th bench and will drop in sixth bench

Answer 2

HEY!!

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▶The horizontal component

▶x = uxt  = ucos53.t.............................................(1)

▶t = x / ucos53

y = uy.t - gt^2 /2 = usin53.t - gt^2 /2........................(2)

substituting the value of t in (2), we get

y = ( u sin 53) *( x / u cos 53 ) -  g ( x / u cos 53 )2  / 2

 or, y = x tan 53  ( gx^2  ) / 2 u^2  cos^2 (53).........................(3)

If we assume the ball strikes the nth step, then

y = net vertical height or distance =  number of steps x height of one step (1m) - height of the barrier (reference)

y = nx1 - 1 m = (n-1) meter

x = distance from player to first bench + additional distance

x =  100 + y = 100 + (n - 1) meter

so, (3) becomes

(n - 1) = (100 + n).(4/3) -  [(4.9x(110 + n)^2 ) /(35 x 35 x .6^2 )]

Or we get the following quadratic equation

n^2 + 190n -1190 = 0

here roots 

n = [-190 +(190^2 - (4x1x-1190))^1/2] / 2

n = (-190 + 202) / 2

n = 6