Question

answer me 36th plz quick

Answer 1

Hey there !!!!

An oblique projectile which is of the form Y=Ax-Bx² and range =A/B

Now comparing y=10x-20x² with Y=Ax-Bx² 
         A=10 B=20

Range = A/B= 10/20 =1/2=0.5m

Hope this helped you !!!!

Answer 2

Hi friend,


Given:

y = 10x− 20x^2

To maximise y ,

we diffrentiate it wrt "x"

dy/dx = 10−2 ×(20) x= 0

x = 10/2×(20)

..................................

y= 10(10/40) −20( 10/40)^2

y=(10)^2/2×(20) −( (10)^2/4(20) )

y=(10)^2/2 (20) −( (10)^2/4(20 )

y= (10)^2/4(20)


Now for the angle just put:

dy/dx= (10)−2(20)x

put x=0;

dy/dx= a=Tanθ

θ=Tan−1a


θ= Tan -1(10)

=Tan -10

=1/2

=0.5

Hope this helps you..☺☺