Question

The equation of motion of a projectile is 3 2 y 12x x 4   The horizontal component of velocity is 3 ms-1. What is the range of the projectile ?

Answer 1

(b) Given y=12x−34x2,ux=3ms−1

vy=dydt=12dxdt−32xdxdt

At x=0,vy=uy=12dxdt=12ux=12×3=36ms−1

ay=ddt(dydt)=12d2xdt2−32[(dxdt)2+xd2xdt2]

But d2xdt2=ax=0. Hence

ay=−32(dxdt)2=−32u2x=−32×(3)2=−272ms−2

Range, R=2uxuyay=2×3×3627/2=16m

Alternatively : We have y=12x−34x2. When projectile again comes to ground, y=0andx=R.

0=12R−34R2⇒R=16m.

Answer 2

Answer:

Explanation:

y = 12x - 3/4x²

the range of the projectile ----  is when the particle reaches the maximum horizontal distance ie y=0

12 x = 3/4x²

x = 16m