Question

Pls....give me the ans......


Trajectory of a body in a projectile motion is given by y= x-(x^2/80).x and y are in meters. Find maximum height of this projectile.

Answer 1

the standard equation of parabola is,

y=(tan α)x - (g/2(ucosα)^2)x^2.....(1)

Given,

y=x - x^2/80..............(2)

Comparing (1) & (2)

tan α=1

α=π/4

g/2(ucos α)^2 = 1/80

10/2(u*1/√2)^2 = 1/80

u^2 = 80*10 =800

Height(H) = (u^2 sin^2 α)/2g

=(800*(1/√2)^2)/2*10

=20 m

Answer 2

MAX HEIGHT IS 20 metres

Given:

• y = x - x²/80

To find:

• Max height ?

Calculation:

First of all, we know that range is obtained when y displacement is zero (i.e. y = 0)

$y = x - \dfrac{ {x}^{2} }{80}$

• Putting y = 0

$\implies \: 0 = x - \dfrac{ {x}^{2} }{80}$

$\implies \: x = \dfrac{ {x}^{2} }{80}$

$\implies \: x = 80 \: m$

• So, range is 80 metres.

Now, for half value of range, the max height is obtained.

So, putting x = 80/2 = 40 m in equation:

$y = x - \dfrac{ {x}^{2} }{80}$

$\implies y = 40 - \dfrac{ {40}^{2} }{80}$

$\implies y = 40 - \dfrac{ 1600 }{80}$

$\implies y = 40 - 20$

$\implies y = 20 \: m$

So, max height is 20 metres.