Physics, 31.08.2019 18:30 omojay1257

Arock is thrown upward from the level ground in such a way that the maximum height of its flight is equal to its horizontal range r.
at what angle is the rock thrown?

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Answer from: miyah5316

Newton's second laws tells us that the vertical motion of the rock in function of the time

follows $y(t)=-\frac12gt^2+v_0t\cdot\sin(\alpha)$ while the horizontal motion follows $x(t)=v_0t\cdot\cos(\alpha)$ where $\alpha$ is the angle of the throw with respect to the ground and v_0

is the initial speed of the rock.

Let's find the horizontal range

: the rock hits the ground when y(t)=0

which happens for $t_0=\dfrac{2v_0\sin(\alpha)}g$ , and it has thus traveled $R=x(t_0)=2\dfrac{v_0^2}g\sin(\alpha)\cos(\alpha)$ .

Let's find the maximum height traveled : the maximum height is obtained when the rock's vertical speed is zero, i.e. when $v_0\sin(\alpha)-gt=0$ that is to say at $t_1=\dfrac{v_0\sin(\alpha)}g$ . At this time it has a height of $y(t_1)=\dfrac{v_0^2\sin^2(\alpha)}{2g}$

Therefore the problem is resolved when $2\dfrac{v_0^2}g\sin(\alpha)\cos(\alpha)=\dfrac{v_0^2\sin^2(\alpha)}{2g}$ i.e. for $\boxed{\alpha=\arctan(4)\approx76^\circ}$

Answer from: sarahidan

The rock is thrown at an angle 76 degrees

Explanation:

The rock thrown will follow projectile motion with

Range , $R=\frac{u^{2}\sin 2\Theta }{g}$

and maximum height , $H_{max}=\frac{u^{2}\sin ^{2}\Theta }{2g}$

where u=initial speed of rock , g= acceleration due to gravity ,

$\Theta = initial\, angle \, of \, through\, with\, horizonatal$

Given $R=H_{max}$

=> $\frac{u^{2}\sin 2\Theta }{g}=\frac{u^{2}\sin ^{2}\Theta }{2g}=2\sin \Theta \cos \Theta =\frac{\sin ^{2}\Theta }{2}$

=> $\tan \Theta =4=\Theta =\tan^{-1}(4)= 76^{\circ}$

Thus the rock is thrown at an angle 76 degrees.

Answer from: Quest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

Answer from: Quest

Domain eukarya; kingdom plantae should be the answer.

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Arock is thrown upward from the level ground in such a way that the maximum height of its flight is...