Question

B. The equation of trajectory of a particle moving in
x - y plane under the gravity is given as
2y = 4x – 8x2. The maximum height reached by the
particle in y-direction will be (where x and y is in
meter)
(1) 1/2 m
(2) 1 m
(3)1/8 m
(4)1/4m Please answer with workings ​

Answer 1

Equation of trajectory

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

Comparing the equations,

tanA=2

And g/2u^2cos^2A=4

u^2=50/8

Place these values in maximum height formula

H=u^2sin^2A/2g

=1/4 m

This is the final answer

I have assumed you know how to find sin and cos from tanA. Also place the value of g as 10m/s^2

If you have anymore questions, please ask..