PHYSICS COLLEGE

A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.When it reaches the highest point of its trajectory, its speed is150 m/s. In a second trial with the same projectile, the initialspeed is the same but the angle is now 37 degree with thehorizontal. At its highest point in this trajectory, the velocityof the projectile would be what?

Answers

Answer 1

Answer:

Answer:

v₂ = 176.24 m/s

Explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

$H_(max)= (v^2 sin^2(\theta))/(g)$

now,

$H_(max)= (v_1^2 sin^2(\theta_1))/(g)$

$H_(max)= (v_2^2 sin^2(\theta_2))/(g)$

now, equating both the equations

$(v_2^2)/(v_1^2)=(sin^2(\theta_1))/(sin^2(\theta_2))$

$(v_2^2)/(150^2)=(sin^2(45^0))/(sin^2(37^0))$

v₂² = 31061.79

v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s

Related Questions

What is the speed of an electron with a de Broglie wavelength of 0.20 nm?
A force of 140 140 newtons is required to hold a spring that has been stretched from its natural length of 40 cm to a length of 60 cm. Find the work done in stretching the spring from 60 cm to 65 cm. First, setup an integral and find a a, b b, and f ( x ) f(x) which would compute the amount of work done.
A toy car is tied to a string and pulled across a table horizontally. Which is thecorrect free-body diagram for this situation?TFNFNTFNENTWWWwАBСDΟ Α. Α
The data listed below are for mechanical waves. Which wave has the greatest energy?
A long, thin superconducting wire carrying a 17 A current passes through the center of a thin, 3.0-cm-diameter ring. A uniform electric field of increasing strength also passes through the ring, parallel to the wire. The magnetic field through the ring is zero.a. At what rate is the electric field strength increasing? b. is the electric field in the direction of the current or opposite to the current?

A 72.0-kg person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface of the door. The doorknob is located 0.800 m from axis of the frictionless hinges of the door. The door begins to rotate with an angular acceleration of 2.00 rad/s 2 . What is the moment of inertia of the door about the hinges?

Answers

Answer:

Moment of inertia will be $I=2kgm^2$

Explanation:

We have given mass of the person m = 72 kg

Radius r = 0.8 m

Force is given F = 5 N

Angular acceleration $\alpha =2rad/sec^2$

Torque is given by $\tau =F* r=5* 0.8=4N-m$

We know that torque is also given by

$\tau =I\alpha$ , here I is moment of inertia and $\alpha$ is angular acceleration

So $4=I* 2$

$I=2kgm^2$

On December 26, 2004, a great earthquake occurred off the coast of Sumatra and triggered immense waves (tsunami) that killed some 200000 people. Satellites observing these waves from space measured 800 from one wave crest to the next and a period between waves of 1.0 hour.Part AWhat was the speed of these waves in m/s?Express your answer using two significant figures.=Part BWhat was the speed of these waves in km/h ?Express your answer using two significant figures.=

Answers

Answers:

a) 222.22 m/s

b) 800.00 km/h

Explanation:

The speed of a wave is given by the following equation:

$v=f \lambda$

Where:

$v$ is the speed

$f=(1)/(T)$ is the frequency, which has an inverse relation with the period $T=1 h$

$\lambda=800 km$ is the wavelength

Solving with the given units:

$v=(1)/(T)\lambda$

$v=(1)/(1 h)800 km$

$v=800.00 km/h$ This is the speed of the wave in km/h

Transforming this speed to m/s:

$v=800.00 (km)/(h) (1 h)/(3600 s) (1000 m)/(1 km)$

$v=222.22 m/s$ This is the speed of the wave in m/s

Anatomy of a Wave worksheet can someone help me out with the answers????

Answers

Part 1: here are the answers in order
5
2
1
3
4

Light from a sodium lamp (λ = 589 nm) illuminates two narrow slits. The fringe spacing on a screen 150 cm behind the slits is 4.0 mm. What is the spacing (in mm) between the two slits?

Answers

The spacing between the two slits is 0.221mm.

The spacing between the two slits is given as,

$D=(\lambda L)/(y)$

Where $\lambda$ is wavelength, y is fringe spacing and L is length of screen.

Given that, $\lambda=589nm,L=150cm,y=4mm$

Substitute in above equation.

$D=(589*10^(-9)*150*10^(-2) )/(4*10^(-3) )\n \nD=2.21*10^(-4) m\n\nD=0.221mm$

Hence, the spacing between the two slits is 0.221mm.

Learn more about the sodium lamp here:

brainly.com/question/24867424

A 1200-kg cannon suddenly fires a 100-kg cannonball at 35 m/s. what is the recoil speed of the cannon? assume that frictional forces are negligible and the cannon is fired horizontally.

Answers

Answer:

Recoil velocity of cannon = 2.92 m/s

Explanation:

By law of conservation of momentum, we have momentum of cannon = momentum of cannonball.

Mass of cannon = 1200 kg

Mass of cannon ball = 100 kg

Velocity of cannon ball = 35 m/s

We have, Momentum of cannon = momentum of cannon ball

1200 x v = 100 x 35

v =3500/1200 = 2.92 m/s

Recoil velocity of cannon = 2.92 m/s

Final answer:

The recoil speed of the cannon is 2.92 m/s.

Explanation:

To find the recoil speed of the cannon, we can use the conservation of momentum. The initial momentum of the cannon and cannonball system is zero since the cannon is at rest before firing. The final momentum is the sum of the momenta of the cannon and cannonball after firing. Using the equation:

Initial momentum = Final momentum

(mass of cannon) x (recoil speed of cannon) = (mass of cannonball) x (velocity of cannonball)

Plugging in the given values:

(1200 kg) x (recoil speed of cannon) = (100 kg) x (35 m/s)

Solving for the recoil speed of the cannon:

recoil speed of cannon = (100 kg x 35 m/s) / 1200 kg = 2.92 m/s

Learn more about recoil speed of cannon here:

brainly.com/question/20887742

#SPJ3

n a distant solar system, a planet of mass 5.0 x 1024 kg orbits a sun of mass 3.0 x 1030 kg at a constant distance of 2.0 x1011 m. How many earth days does it take for the planet ot execute one complete orbit about the sun

Answers

Answer:

F = M2 ω^2 R centripetal force of sun on planet

ω = (F / (M2 R))^1/2 = 2 pi f = 2 pi / P where P is the period

P = 2 pi (M2 * R / F)^1/2

F = G M1 M2 / R^2 gravitational force on planet

P = 2 pi {R^3 / (G M1)]^1/2

P = 6.28 [(2.0E11)^3 / (6.67E-11 * 3.0E30)]^1/2

P = 6.28 (8 / 20)^1/2 E7 = 3.9E7 sec

1 yr = 3600 * 24 * 365 = 3.15E7 sec

P = 3.9 / 3.2 = 1.2 years

Other Questions

Those who support the alien theory believe that the ancient Egyptians where very primitive. What evidence about the ancient Egyptians suggests that claim is inaccurate?

X-rays have wavelengths between 0.1 to 10 nanometers (x10-9). What is the range of its frequency? 3x1017-3x1015 Hz 3x10, 18, -3x10, 16, Hz 3x1018-3x1017 Hz 3x1017-3x1016 Hz

Which statement best describes how the first quatrain relates to the second quatrain? The first shows the beloved’s actions; the second describes how she imitates them. Both the first and the second show the actions of the speaker and the beloved. The first shows the speaker’s actions; the second shows the beloved’s opposition to them. The first shows the speaker’s sadness; the second shows the beloved’s anger.

Assume the following values: d1 = 0.880 m , d2 = 1.11 m , d3 = 0.560 m , d4 = 2.08 m , F1 = 510 N , F2 = 306 N , F3 = 501 N , F4 = 407 N , and MA = 1504 N⋅m . Express the Cartesian components of the resultant force and the couple moment in newtons and newton-meters to three significant figures separated by commas.