The benches of a gallery of a cricket stadium are 1m high and 1m wide.A batsman strikes the ball at a level 1m above the ground and hits a mammoth sixer.The ball starts at 35 m/s at an angle of 53 degrees with the horizontal.The benches are perpendicular to the plane of motion and the first bench is 110m from the batsman.On which bench will the ball hit?

Range of projectile = u².sin 2x /g

                            = 117.75 mts

so on the ground ,it covers 110 mts and the left distance of 7.75 mts is on gallary

each bench hz. length is 1 mts

so it lands on 8th bench

--

Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult
to understand - post it here and we will get you the answer and detailed solution very quickly.We are all
IITians and here to help you in your IIT JEE preparation. All the best.

Regards,
Ramesh
IIT Kgp - 05 batch

The answer to this is 6th bench.

y=(n-1)1 (because there is a 1m uplevel).

x= 110 + (n-1)

x= 110+ y

Put it in the equation of trajectory = 

y = xtanA - gx^2sec^2A/2u^2

where x=110+y

sec62a = 25/9 and u = 35m/s

you will get y =5

put it in y=n-1

we will get n=6

range= (2* u  sinA * u sinB)/g

u sinA= 35 * 4/5= 28

u sinB= 35* 3/5= 21

g=10 m/s² 

range= 2*28*21/10

=42*28/10

=112.6

1st bench 110 to 111

2nd bench 111 to 112

3rd bench 112 to 113

hence, 3rd bench.