Question

the horizontal and vertical displacement of the projectile has given as x is equal to at and Y is equal to b t minus Ct
square then the velocity of the projection is​

Answer 1

Answer:

$\vec{v}=v_x\vec{i}+v_y\vec{i}=a\vec{i}+(b-2ct)\vec{j}$

Explanation:

1) We have given that,

Horizontal displacement of projectile,  

$x=at$

Vertical displacement of projectile,

$y=bt-ct^2$

where, ab,c are constants

2)

Horizontal velocity of projectile,  

$v_x=\frac{dx}{dt}=\frac{d(at)}{dt}=a$

Vertical velocity of projectile,  

$v_y=\frac{dy}{dt}=\frac{d(bt-ct^2)}{dt}=b-2ct$

3) Velocity of projection ,

$\vec{v}=v_x\vec{i}+v_y\vec{i}=a\vec{i}+(b-2ct)\vec{j}$

Magnitude of velocity,

$|v|=\sqrt{v_x^2+v_y^2}=\sqrt{a^2+(b-2ct)^2}=\sqrt{a^2+b^2-4bct+4c^2t^2}$

Answer 2

Answer:

$v=\sqrt{a^2+b^2+4C^2t^2-4bCt}$

Explanation:

It is given that,

Horizontal component of the projectile,

$x=at$.............(1)

Vertical component of the projectile,

$y=bt-Ct^2$..............(2)

Differentiating equation (1) wrt t as :

$\dfrac{dx}{dt}=v_x=a$

Differentiating equation (2) wrt t as :

$\dfrac{dy}{dt}=v_y=b-2Ct$

Let v is the velocity of projection. It is given by :

$v=\sqrt{v_x^2+v_y^2}$

$v=\sqrt{a^2+(b-2Ct)^2}$

$v=\sqrt{a^2+b^2+4C^2t^2-4bCt}$

So, the velocity of the projection is $\sqrt{a^2+b^2+4C^2t^2-4bCt}$. Hence, this is the required solution.