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A robot probe drops a camera off the rim of a 239 m high cliff on mars, where the free-fall acceleration is −3.7 m/s2 .
a. find the velocity with which the camera hits the ground.
b. find the time required for it to hit the ground.

Respuesta :

Kalahira
a. We can find the velocity when the camera hits the ground. v^2 = (v0)^2 + 2ay = 0 + 2ay v = sqrt{ 2ay } v = sqrt{ (2)(3.7 m/s^2)(239 m) } v = 42 m/s The camera hits the ground with a velocity of 42 m/s b. We can find the time it takes for the camera to hit the ground. y = (1/2) a t^2 t^2 = 2y / a t = sqrt{ 2y / a } t = sqrt{ (2)(239 m) / 3.7 m/s^2 } t = 11.4 seconds
       
It takes 11.4 seconds for the camera to hit the ground.
whitneytr12

For the camera that is dropped by a robot on a high cliff on mars, we have:

a. The velocity with which the camera hits the ground is 42.05 m/s.

b. The time required for it to hit the ground is 11.4 s.

a. The velocity when the camera hits the ground can be calculated with the following equation:

[tex] v_{f}^{2} = v_{0}^{2} - 2gh [/tex]

Where:

[tex] v_{f}[/tex]: is the final speed =?

[tex] v_{0}[/tex]: is the initial speed = 0 (the camera is dropped)

g: is the acceleration due to gravity in mars = -3.7 m/s²

h: is the height = 239 m

Hence, the final speed is:

[tex] v_{f} = \sqrt{2gh} = \sqrt{2*3.7 m/s^{2}*239 m} = 42.05 m/s [/tex]

b. The time required for it to hit the ground is the following:

[tex] t = -\frac{v_{f} - v_{0}}{g} = -\frac{42.05 m/s}{-3.7 m/s^{2}} = 11.4 s [/tex]

Therefore, the camera hit the ground after 11.4 seconds.

You can see another example of free fall here: https://brainly.com/question/17436247?referrer=searchResults              

I hope it helps you!

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