Name: Motion With Variable Acceleration
Uploaded: 2019-07-19
Duration: 10 min 54 s
Description: You may be aware of the SUVAT equation, but do you think this can be applicable for all types of motion? No, what are those motion? Watch this to find out.

Question

The equation of a projectile is

y = \sqrt{3} x - \frac{g x^{2}}{2},

find the angle of projection. Also, find the speed of projection where

t = 0

,

y = 0

and

x = 0

also

\frac{d^{2} x}{d t^{2}} = 0

and

\frac{d^{2} y}{d t^{2}} = - g

.

Accepted Answer

Given equation of a projectile is $y = \sqrt{3} x - \frac{g x^{2}}{2}$ .

Differentiating w.r.t time, we get

$\frac{d y}{d t} = \sqrt{3} \frac{d x}{d t} - \frac{2 g x \frac{d x}{d t}}{2} = \sqrt{3} \frac{d x}{d t} - g x \frac{d x}{d t}$ .

We know that $\frac{d y}{d t}$ is the vertical velocity $(v_{y})$ and $\frac{d x}{d t}$ is the horizontal velocity $(v_{x})$ . At $t = 0$ and $x = 0$ ,

$\frac{d y}{d t} = \sqrt{3} \frac{d x}{d t} - g (0) \frac{d x}{d t} = \sqrt{3} \frac{d x}{d t} \Rightarrow v_{y} = \sqrt{3} v_{x}$ .

We know that $\tan (θ) = \frac{v_{y}}{v_{x}}$ , where $θ$ is the angle of projection. This gives us $\frac{d^{2} y}{d t^{2}} = - g$ .

Differentiating the first result again w.r.t time we get

$\frac{d^{2} y}{d t^{2}} = \sqrt{3} \frac{d^{2} x}{d t^{2}} - g {(\frac{d x}{d t})}^{2} - g x \frac{d^{2} x}{d t^{2}}$ .

At $t = 0$ , and $x = 0$ this gives us,

$- g = \sqrt{3} \times 0 - g {(v_{x})}^{2} - g \times 0 \times 0$ (Here, $\frac{d^{2} y}{d t^{2}} = - g$ )

$\Rightarrow {v_{x}}^{2} = 1 \Rightarrow v_{x} = 1 m s^{- 1}$

Vertical velocity at this time is $v_{y} = \sqrt{3} v_{x} = \sqrt{3} m s^{- 1}$ .

Speed of projection is the resultant of the horizontal and vertical velocity, i.e., $\sqrt{{v_{x}}^{2} + {v_{y}}^{2}} = \sqrt{1^{2} + {(\sqrt{3})}^{2}} = 2 m s^{- 1}$ .

The equation of a projectile is $y = \sqrt{3} x - \frac{g x^{2}}{2},$ find the angle of projection. Also, find the speed of projection where $t = 0$ , $y = 0$ and $x = 0$ also $\frac{d^{2} x}{d t^{2}} = 0$ and $\frac{d^{2} y}{d t^{2}} = - g$ .

Important Questions on Kinematics

Learn and Prepare for any exam you want !

The equation of a projectile is y=3x-gx22, find the angle of projection. Also, find the speed of projection where t=0 , y=0 and x=0 also d2xdt2=0 and d2ydt2=-g.

Important Questions on Kinematics

Learn and Prepare for any exam you want !

The equation of a projectile is $y = \sqrt{3} x - \frac{g x^{2}}{2},$ find the angle of projection. Also, find the speed of projection where $t = 0$ , $y = 0$ and $x = 0$ also $\frac{d^{2} x}{d t^{2}} = 0$ and $\frac{d^{2} y}{d t^{2}} = - g$ .