All categories

3 years ago

If a ball is 10m high with what velocity will it fall?

Physics

1 answer:

Semmy [17]3 years ago

6 0

14m/s

Explanation:

Given parameters:

Height of the ball = 10m

Unknown:

Velocity of fall or final velocity = ?

Solution:

We are going to use the appropriate equation of motion to solve this problem.

The object is falling with respect to gravity.

V² = U² + 2gH

where V is the final velocity

U is the initial velocity

g is the acceleration due to gravity 9.8m/s²

H is the height of fall

The initial velocity here is zero and

V² = 2 x 9.8 x 10 = 196

V = 14m/s

learn more:

Motion problems brainly.com/question/5248528

#learnwithBrainly

You might be interested in

1.9 km converted to miles

Snowcat [4.5K]

When you convet km to miles this is what you get

1.18061

7 0

3 years ago

Compared to a stationary galaxy, light from a galaxy that is moving away from earth will appear _____.

Vlad1618 [11]

"B" When an object moves away from us, the light is shifted to the red end of the spectrum, as its wavelengths get longer.

3 0

3 years ago

Read 2 more answers

Forces can act on an object in the same direction or in opposite. how does each situation affect the motion of the object?

Anon25 [30]

Hi pupil here's your answer ::

➡➡➡➡➡➡➡➡➡➡➡➡➡

Action and Reaction do not act on the same body !! If they acted on the same body, the resultant force will be zero and their could be never accelerated motion.

If both the forces acted on the same body, then if they are equal to opposite direction the object will remain stationary. If on of the forces is greater than other the object will move in the direction of greater force.

If both acted in the same direction there would be an accelrated motion.

⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅

Hope this helps . . . . .

5 0

3 years ago

Se lanza verticalmente una esfera con una rapidez de 30m/se. Determinar la rapidez de la esfera a una altura de 40m (g=10m/s2)

sergeinik [125]

$v^2-{v_0}^2=2a(x-x_0)$

dónde $v$ es la velocidad de la esfera, $v_0$ es suya velocidad inicial, $a=-g$ la aceleración debida a la gravedad, $x$ la posición, y $x_0$ la posición inicial. Tomamos $x_0=0\,\mathrm m$ a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

Entonces

$v^2-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-10\,\dfrac{\mathrm m}{\mathrm s^2}\right)(40\,\mathrm m)$

$\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}$

Esto nos dice que la esfera alcanza una altura de 40 m en dos momentos - una vez hacia arriba y una vez hacia abajo. Sin embargo, independientemente del signo de la velocidad, sabemos que suya magnitud es 10 m/s, y así tenemos una rapidez de 10 m/s también en ambos momentos.

4 0

3 years ago

Ship A is located 4.1 km north and 2.3 km east of ship B. Ship A has a velocity of 22 km/h toward the south and Ship B has a vel

castortr0y [4]

Answer:

a) x component = -31.25 km/hr

b) y component = 46.64 km/hr

Explanation:

Given data:

A position is 4km north and 2.5 km east to B

Ship A velocity = 22 km/hr

ship B velocity = 40 km/hr