# 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. )

Abhishek Doyal
8 years ago
Q2 please tell me the width of river
Abhishek doyal
IIT Bombay
Abhishek Doyal
8 years ago
Abhishek doyal
IIT bombay
sunayanaa
8 Points
8 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
8 Points
8 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)
Arun Kumar IIT Delhi
8 years ago
Hello
$\\we will answer first question only for the second \\make another post \\20=v_y4-5*4^2 \\5=v_y-20 \\=>v_y=25m/s \\=>50sin\theta=25 \\=>sin\theta=1/2$
Thanks & Regards
Arun Kumar
Btech, IIT Delhi