" f the equation of motion of a proje-ctiel is "y=3x-(1)/(8)x^(2)," the range and maximum height are respectively "(y" and "x" are in metres) "

The equation of a trajectory of a projectile is y=8x-4x^(2) . The maximum height of the projectile is ?

The equation of motion of a projectile is y = 12 x - (3)/(4) x^2 . The horizontal component of velocity is 3 ms^-1 . What is the range of the projectile ?

The equation of trajectory of a projectile is y=10x-(5/9)x^2 if we assume g=10ms^(-3) then range of projectile (in metre) is

A particle travels according to the equation y=x-(x^(2)/(2)) . Find the maximum height it acheives (x and y are in metre)

If (x-y,x+y)=(2,8), then the values of x and y are respectively.

If x+y=60 and xy^(3) is maximum, then x and y respectively will be

Solve the equations for x and y3^(x)=9x3^(y),8x2^(y)=4^(x)

The equation y=4+2 sin (6t-3x) represents a wave motion. Then, wave speed and amplitude, respectively are