Question

The equation of trajectory of a projectile thrown
on the ground is 5x^2 - 2x + 2y = 0. Here x, y
are in meter and denote the horizontal and
vertical distance respectively. The point of
projection is to be assumed as origin. Find out
(i) horizontal range
(ii) maximum height allained by the projectile
(iii) angle of projection
(iv) time of ſight
(v) velocity of projection (g = 10 ms)​

Answer 1

The point of projection is (0,0)

Given,

Equation of trajectory : 5x^2 - 2x + 2y = 0

Horizontal Range :

The distance between the point of projection and the point where the body meets ground again.

When the body completes its flight and lands, The Displacement is entirely in x axis.

So for Range R, y = 0.

From equation of Trajectory

⇒ 5x^2 - 2x + 2y = 0

⇒ 5x² - 2x + 0 = 0

⇒ x ( 5x - 2) = 0

⇒ 5x - 2 = 0, x ≠ 0

⇒x = 2/5 m

⇒ R = 2/5 = 0.4 m.

Maximum height attained by the body

The maximum height is the vertical displacement at the peak of the trajectory.

For Maximum height, Vertical velocity = 0

From equation of Trajectory,

⇒ 5x^2 - 2x + 2y = 0

Differentiating with respect to t, gives

⇒ 10x dx/dt - 2dx/dt + 2dy/dt = 0

⇒ (5x-1)dx/dt + 2(dy/dt) =0

Since dy/dt represents Vertical velocity, dy/dt = 0

⇒ (5x-1) dx/dt = 0

Since there is no retarding force in horizontal direction, dx/dt ≠ 0

⇒ 5x - 1 = 0

⇒ x = 1/5

So corresponding value of y gives the maximum height

⇒ 5x^2 - 2x + 2y = 0

⇒ 5(1/5)² - 2(1/5) + 2y = 0

⇒ 1/5 - 2/5 + 2y = 0

⇒ 2y = 1/5

⇒ y = 1/10

So, H = 0.1 m.

Angle of projection

⇒tanθ = 4H/R

⇒ tanθ = 4(0.1) / 0.4

⇒ tanθ = 1

⇒θ = tan^-1 ( 1)

⇒θ = 45°

Velocity of projection

$\displaystyle H = \frac{ {u}^{2} \sin ^{2} ( \theta) }{2g}$

So,

⇒0.1(2)(10) = u² sin²45

⇒2 = u² ( 1/2)

⇒u² = 4

⇒u = 2 m/s

Time of flight

$\displaystyle\:t\:\:=\frac{2u \sin \theta}{g}$

So,

⇒ t = 2(2)(1/√2) / 10

⇒ t = 2√2 / 10

⇒ t = 0.28 seconds.