Question

the equation of trajectory of a projectile is given as y=2x–x²/2 the maximum height of projectile is​

Answer 1

answer : 2m

explanation : The equation of trajectory of a projectile is given as $y=2x-\frac{x^2}{2}$

differentiating y with respect to x,

dy/dx = d(2x - x²/2)/dx

= 2 - 2x/2

= 2 - x

now, find value of x at dy/dx = 0

so, dy/dx = 2 - x = 0 ⇒x = 2

again differentiate with respect to x,

d²y/dx² = -1 < 0 hence, function will gain maximum value at x = 2

so, put x = 2 in equation y = 2x - x²/2

then, y = 2(2) - (2)²/2 = 4 - 2 = 2m

hence, maximum height of projectile is 2m.

method 2 :

if you are not familiar with math, we have another option too.

you know standard trajectory of projectile motion, y = tanα.x - gx²/2u²cos²α

on comparing with y = 2x - x²/2

we get, tanα = 2 ⇒cosα = 1/√5 and sinα = 2/√5

g/2u²cos²α = 1/2

or, g = u²(1/√5)²

or, 10 × 5 = u² ⇒u² = 50

now, maximum height , H = u²sin²α/2g

= {50 × (2/√5)²}/2(10)

= {50 × 4/5}/20

= 40/20 = 2m

hence, maximum height = 2m

Answer 2

Answer:

2m

Explanation:

y=2x-$x^{2}$/2 is the equation of a parabola.

At maximum height Hmax, the slope of this graph will be zero. i.e,dy/dx=0

and Hmax will equal to the value y at that point.

ATQ, y=2x-$x^{2}$/2

differentiating both sides with respect to x,

dy/dx=2-1/2×2x

o=2-2x           {since dy/dx=0}

x=2.

Now we find the value of y for which x=2 and that gives Hmax.

ATQ, y=2x-$x^{2}$/2

y=2(2)-4/2=4-2=2m