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2 years ago

How far from the castle wall does the launched rock hit the ground?

2 answers:

7 0

Khan Academy has a really good video on this: https://m.youtube.com/watch?v=a5QnXi_lZII&ebc=ANyPxKqwT2wePxBeuoMvGuAgopefPsAAET3fCVrN7oTr7j7jOobG5sJS_Db_czLaVyZJioXM2Pt-yDUZ_tzi1nX6bCAZWTpZUQ

I'll get you started really quick (but it's way past my bedtime, so... ;)).

30 sin(28) = 14.1
30 cos(28) = 26.5

This means, when it was launched, it was launched 14.1 m/s upwards and 26.5 m/s outwards/horizontally. If you recal horizontally-launched projectile motion, there is negligible resistance (we just have to find the force of gravity against the vertical component). Calculate the time it takes by plugging in the vertical displacement into one of your mechanics equations, and then using a range/distance formula for the rest.

(The video is more helpful than I'm being at 2:08 am. Have fun and good luck!)

lina2011 [118]2 years ago

4 0

King Arthur's knights use a catapult to launch a rock from their vantage point on top of the castle wall, 14 m above the moat. The rock is launched at a speed of 27 m/s and an angle of 32degrees above the horizontal.

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True or False:

bulgar [2K]

Answer:

False

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False

8 0

2 years ago

while flying a plane parallel to the ground a pilot releases a fuel tank in order to reduce the planes mass. what is the tanks f

stepan [7]

La velocidad vertical del tanque después de caer 10 m es 14 m/seg .

La velocidad vertical del tanque se calcula mediante la aplicación de la fórmula de velocidad , la componente vertical Vfy, del movimiento horizontal como se muestra a continuación :

Vfy=?

h = 10 m

Fórmula de Velocidad vertical Vfy:

Vfy² = 2*g*h

Vfy= √(2*9.8m/seg2* 10m )

Vfy= 14 m/seg

7 0

2 years ago

Un ladrillo se le imparte una velocidad inicial de 6m/s en su trayectoria hacia abajo. ¿cual sera su velocidad final despues de

marshall27 [118]

Saludos!

Respuesta:

28,64 m/s.

Explicación:

Datos:

Altura o distancia recorrida: 40 m
Vo: 6 m/s
Aceleración de la gravedad: 9,81 m/s²

El ejercicio puede ser resuelto facilmente utilizando la siguiente formula, sin embargo es posible realizarlo utilizando formulas diferentes.

Entonces tenemos que:

$Vf ^{2} -Vo ^{2} =2 x g x h$

Es importante saber que al estar lanzando el ladrillo hacia abajo, el sentido del movimiento sigue el sentido de la gravedad, es decir es necesario que tomes el valor de la gravedad como positivo (+) y no negativo (-) como normalmente se usa.

Sustituyendo tenemos que:

$Vf x^{2} =(6 m/s) ^{2} +(2)x(9,81 m/s ^{2})x(40m) \\ Vf ^{2} =820,8 m^{2} /s ^{2} \\ \sqrt{Vf ^{2} } = \sqrt{820,8m^{2}/s ^{2} } \\ Vf=28,64 m/s$

Que tengas un buen día!

6 0

3 years ago

A girl hits a baseball with a bat. When the bat strikes the ball, what is the

Illusion [34]

Answer:

Its simple! B The ball accelerates upward!♡♡

5 0

3 years ago

Read 2 more answers

The greater the amplitude of a wave a. The quieter the sound b. The higher the frequency of the sound c. The dimmer the light in

Stolb23 [73]

Answer:

The greater the amplitude the greater the energy.

(Think of a water wave - which carries greater energy a 1 ft wave or

a 10 ft wave)

8 0

2 years ago