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Q .1- A projectile projected at some angle with vel. 50m/s crosses a 20m wall after 4s from the time of projection. The angle of projection of particle will be ? ( pls give full sol.) 2. a swimmer can swim in still water at 1.5m/s , the river is flowing with 2m/s from left to right. The minimum time taken by swimmer to cross the river and drift will be? ( pls full sol. )

AJAY KUMAR

12 Years agoGrade 12th pass

5 Answers

Abhishek Doyal

12 Years ago

Q2 please tell me the width of river

Abhishek doyal

IIT Bombay

Askiitian tutor

Abhishek Doyal

12 Years ago

Abhishek doyal

IIT bombay

Askiitian tutor

sunayanaa

12 Years ago

qn 1:

We can solve almost every projectile question using just the basic kinematics formulas.

Let us take the sign convention for this problem as g (acceleration due to gravity) as -10m/s^2 in the upward direction (+y axis) and +10m/s^2 in the downward direction (-y axis).

Let initial vel be U

U=50m/s (given)

Let time be t

t=4 sec (given)

let x be the angle of projection which we have to find.

Here the 20 m wall is the vertical distance covered by the projectile.

The range of the projectile would be the horizontal distance covered by it but we dont need to worry about it here.

We can split ‘U’ into its rectangular components

the vertical displacement 20 m of the projectile is due to only its vertical component of U (Uy) which is Usinx.

=50sinx

y(vertical displacement)=Uyt+1/2(-g)(t)^2

=Uyt-1/2(g)(t)^2

20=50sinx-1/2(10)(4)^2

= 20=50sinx-(5)(16)

= 20=50sinx-80

= 50sinx=100

= sinx=1/2

= x=30 degrees

Hope this helps!!!!

sunayanaa

12 Years ago

Small correction. The procedure and the final answer is the same but i made a calculation mistake in between

i missed the t there

20=50sinx * 4- 5*16

=200sinx-80

100=200sinx

sinx=1/2

hence x=30 degrees

(in my previous calculations we are getting sinx=100/50 =2 which is obviously wrong)

sorry about that

Arun Kumar

12 Years ago

Hello

$\\$we will answer first question only for the second $ \\$make another post$ \\20=v_y4-5*4^2 \\5=v_y-20 \\=>v_y=25m/s \\=>50sin\theta=25 \\=>sin\theta=1/2$