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2 years ago

Topic: Chapter 10: Projectory or trajectile?

Physics

1 answer:

Charra [1.4K]2 years ago

5 0

Answer:

The other angle is 75⁰

Explanation:

Given;

velocity of the projectile, v = 10 m/s

range of the projectile, R = 5.1 m

angle of projection, 15⁰

The range of a projectile is given as;

$R = \frac{u^2sin(2\theta)}{g}$

To find another angle of projection to give the same range;

$5.1 = \frac{10^2 sin(2\theta)}{9.81} \\\\100sin(2\theta) = 50\\\\sin(2\theta) = 0.5\\\\2\theta = sin^{-1}(0.5)\\\\2\theta = 30^0\\\\\theta = 15^0\\\\since \ the \ angle \ occurs \ in \ \ the \ first \ quadrant,\ the \ equivalent \ angle \\ is \ calculated \ as;\\\\90- \theta = 15^0\\\\\theta = 90 - 15^0\\\\\theta = 75^0$

<u>Check: </u>

sin(2θ) = sin(2 x 75) = sin(150) = 0.5

sin(2θ) = sin(2 x 15) = sin(30) = 0.5

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3 years ago

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Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

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(d): $\rm 2c-6bt.$

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Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

For $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = \dfrac{2c}{3b}$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.$

Here,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Thus, the particle reach its maximum x value at time $\rm t_o = \dfrac{2c}{3b}.$

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3 years ago

Can a vector be shorter than one of its components

Sauron [17]

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3 years ago

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kolbaska11 [484]

Answer:

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Explanation:

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K= constant of Coulomb's law = 9 ×10^9 Nm^2C-2

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r= distance of separation

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3 years ago

A bungee cord is 30m long and when stretched a distance x itexerts a restoring force of magnitude kx. Your father-in-law (mass95

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the force is F=380 N

the gravitational acceleration is g=9.8 m/s2

The equation of k is given by equating the energy at the equilibrium point which is given as

$U_{potential}=U_{elastic}\\mgh=\dfrac{1}{2} k y^2\\k=\dfrac{2mgh}{y^2}$

Here

m=95 kg, g=9.8 m/s2, h=41 m, y=11 m so

$k=\dfrac{2mgh}{y^2}\\k=\dfrac{2\times 95\times 9.8\times 41}{11^2}\\k=630.92 N/m$

Now the force is

$F=kx\\$ or

$x=\dfrac{F}{k}\\$

So here F=380 N, k=630.92 N/m

$x=\dfrac{F}{k}\\x=\dfrac{380}{630.92}\\x=0.602 m$

So the distance is 0.602 m

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3 years ago