Question

If the equation of motion of a projectile is 3x-1/8x^2, the range and maximum height are respectively.

Answer 1

As we know
Motion along horizontal direction
is given by this equation :
x = xo + ux t + 1 ax t^2 /2

Now as given
3x- 1 / 8x^2

3x- 1 / 8x^2 = u x t cos theta

u = 3x -1 / 8x ^2 t cos theta

Maximum height :

h = u^2 sin^2 theta / 2g
h = ( 3x- 1 /8x^2 t cos theta ) ^2 sin^2 theta /2g
= (9x^2 + 1 - 6x ) sin^2 / 8x^4 Cos theta t^2 x 2g

Horizontal Range :

R= u^2 sin 2 theta / g
= ( 3x -1 / 8x^2 t cos theta )^2 sin 2 theta / g
= ( 9x^2 + 1 - 6x ) sin 2 theta / 8x^4 Cos theta t^2 x g
__________________Maybe its helpful

Answer 2

Hope this helps you..