Ask question

Ask question

All categories

alukav5142 [94]

2 years ago

13

Solve the problem. Unless stated otherwise, assume that the projectile flight is ideal, that the launch angle is measured from t

he horizontal, and that the projectile is launched from the origin over a horizontal surface An ideal projectile is launched from level ground at a launch angle of 26° and an initial speed of 48 m/sec. How far away from the launch point does the projectile hit the ground

1 answer:

kirill115 [55]2 years ago

3 0

Answer:

d = 185.26 meters

Explanation:

It is given that,

Launching angle of the projectile, $\theta=26^{\circ}$

Initial speed of the projectile, u = 48 m/s

Let at distance d the projectile hits the ground from the launch point. It is equal to range of the projectile. Its formula is given by :

$d=\dfrac{u^2\ sin2\theta}{g}$

Substituting all the values in above formula. So, we get :

$d=\dfrac{(48)^2\ sin2(26)}{9.8}$

d = 185.26 meters

So, the distance between the launch point and the point where it hits is 185.26 meters. Hence, this is the required solution.

You might be interested in

How much heat is evolved when 1234 g of water condenses to a liquid at 100.°c?

lbvjy [14]

During the phase transition vapour --> liquid water, the temperature of the water does not change; the molecules of water release heat and the amounf of heat released is equal to
$Q=mL_e$
where
m is the mass of the water
$L_e$ is the latent heat of evaporation.

For water, the latent heat of evaporation is $L=2257 kJ/kg$ , while the mass of the water is
$m=1234 g=1.234 kg$

so, the amount of heat released in the process is
$Q=mL_e = (1.234 kg)(2257 kJ/kg)=2785 kJ$

6 0

3 years ago

If a roller coaster cart, with a mass of 100 kg, traveled this coaster, how much kinetic energy would it have at point 'E'?

zzz [600]

Answer:

Explanation:

Assuming no friction between the roller coaster car and the hill, and neglecting air resistance, the kinetic energy the roller coaster car would have at the bottom of the hill would be equal to its gravitational potential energy at the top of the hill, by conservation of energy.

8 0

3 years ago

How could you increase the sleds acceleration

gladu [14]

By putting this special transportation plastic on the bottom of the sled, because the transportation plastic is slick. that is what the transport bins slide around on. (p.s) the plastic is really expensive!

6 0

2 years ago

A 2.93 kg particle has a velocity of (2.98 i hat - 3.98 j) m/s.

cupoosta [38]

Answer:

a) The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

Explanation:

a) The vectorial equation of momentum is represented by the following expression:

$\vec p = m\cdot \vec v$ (1)

Where:

$\vec p$ - Vector momentum, measured in kilogram-meters per second.

$m$ - Mass of the particle, measured in kilograms.

$\vec v$ - Vector velocity, measured in meters per second.

If we know that $m = 2.93\,kg$ and $\vec v = 2.98\,\hat{i}-3.98\,\hat{j}\,\,\,\left[\frac{m}{s} \right]$ , then the momentum is:

$\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right]$

$\vec p = 8.731\,\hat{i}-11.661\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]$

The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of momentum are represented by the following expressions:

$\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}}$ (2)

$\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right)$ (3)

Where:

$\|\vec p\|$ - Magnitude of momentum, measured in kilogram-meters per second.

$\theta$ - Direction of momentum, measured in sexagesimal degrees.

If we know that $p_{x} = 8.731\,\frac{kg\cdot m}{s}$ and $p_{y} = -11.661\,\frac{kg\cdot m}{s}$ , then the magnitude and direction of momentum are, respectively:

$\|\vec p\| = \sqrt{\left(8.731\,\frac{kg\cdot m}{s} \right)^{2}+\left(-11.661\,\frac{kg\cdot m}{s} \right)^{2}}$

$\|\vec p\| \approx 14.567\,\frac{kg\cdot m}{s}$

$\theta =\tan^{-1}\left(\frac{-11.661\,\frac{kg\cdot m}{s} }{8.731\,\frac{kg\cdot m}{s} } \right)$

$\theta \approx 306.823^{\circ}$

The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

6 0

3 years ago

A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second, and the t

slega [8]

Answer:

The tension increases to four times its original value.

Explanation:

v = Velocity

r = Radius

m = Mass of stone

The centripetal force is

$F_c=m\dfrac{v^2}{r}$

The tension will balance the centripetal force

$T=m\dfrac{v^2}{r}$

$\\\Rightarrow T\propto v^2$

$\dfrac{T_1}{T_2}=\dfrac{v_1^2}{v_2^2}\\\Rightarrow \dfrac{T_1}{T_2}=\dfrac{v_1}{2^2v_1^2}\\\Rightarrow \dfrac{T_1}{T_2}=\dfrac{1}{4}\\\Rightarrow T_2=4T_1$

The new tension will be 4 times the old tension

5 0

2 years ago