Question

The path of projectile is defined by equation t equals to 3t minus t square upon 30 and theta square equals to 1600 minus t square . find velocity after 30 seconds

Answer 1

Answer:

Hi

Explanation:

Try to make the question clear

Answer 2

The velocity of the projectile after 30 seconds is $r - \frac{180}{\sqrt{7} }$ × θ.

Given:

The equations for the path of the projectile = $3t - \frac{1}{30} t^2$ and θ² = $1600 - t^2$.

The time, t = 30 seconds.

To Find:

We have to find the velocity of the projectile after 30 seconds.

Solution:

The expression for the velocity of the particle that  moves along a curve in a plane is given by,

$v = \frac{dr}{dt}. r+ r. \frac{dθ}{dt}.θ$

Given that, $r = 3t - \frac{1}{30} t^2$.

On differentiating the above equation, we get,

∴, $\frac{dr}{dt} = 3 - \frac{1}{30}$ × $2t$

Or,  $\frac{dr}{dt} = 3 - \frac{1}{15}t$.

At t = 30s, the values of r and $\frac{dr}{dt}$ becomes,

$r = 3 . (30) - \frac{1}{30} .(30)^2 = 90 - 30 = 60 m.$

$\frac{dr}{dt} = 3 - \frac{1}{15}. (30) = 1 m/s.$

Given that, θ² = $1600 - t^2$.

∴, θ = $\sqrt{1600 - t^2}$.

On differentiating the above equation, we get,

$\frac{dθ}{dt} = \frac{d}{dt} (\sqrt{1600 - t^2}) = \frac{(-2t)}{2.[\sqrt{1600 - t^2}]}$

On simplifying, the above equation becomes,

$\frac{dθ}{dt} = \frac{t}{\sqrt{1600 - t^2}}$.

At t = 30s, the values of θ and $\frac{dθ}{dt}$ becomes,

θ = $\sqrt{1600 - (30)^2} = \sqrt{700} = 10\sqrt{7} rad.$

$\frac{dθ}{dt} = \frac{30}{\sqrt{1600 - (30)^2}} = - \frac{30}{10\sqrt{7} } = - \frac{3}{\sqrt{7} } rad/s.$

On substituting these values of r, $\frac{dr}{dt}$, θ, and $\frac{dθ}{dt}$ in the expression for velocity of the particle that  moves along a curve in a plane, we get,

$v = 1$ × $r+ 60$ × $- (\frac{3}{\sqrt{7} })$ × θ

$i.e., v = r - \frac{180}{\sqrt{7} }$ × θ.

Hence, the velocity of the projectile after 30 seconds is $r - \frac{180}{\sqrt{7} }$ × θ.

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