Question

A robot probe on Mars drops a camera off a cliff. On Mars the acceleration of gravity is 3.7 m/s2. The camera is moving down at a speed of 61.0 m/s when it strikes the ground.

a. Find the height from which the camera fell.
b. Find the velocity with which it hits the ground.

Answer 1

The camera fell from a height of approximately 240.56 m on Mars. The velocity with which it hits the ground is 61.0 m/s.

Step-by-step explanation:

To find the height from which the camera fell, we can use the equation for gravitational potential energy: PE = mgh, where m is the mass of the camera, g is the acceleration due to gravity, and h is the height. In this case, the camera's mass is not given, but we can use its weight (mg) instead. Therefore, the equation becomes: PE = wh, where w is the weight of the camera. Rearranging the equation to solve for h, we have: h = PE / w. The initial potential energy (PEi) of the camera is zero, since it is not moving. The final potential energy (PEf) is given by: PEf = whf, where hf is the height from which the camera fell. Substituting the values into the equation, we have: PEf = (10 N)(hf) = (10 N)(9.8 m/s^2)(hf) = 98 hf J. Since the camera was moving downwards, its final velocity (vf) is given. We can use the equation: vf^2 = vi^2 + 2ad, where vf is the final velocity, vi is the initial velocity (0 m/s in this case), a is the acceleration due to gravity (-3.7 m/s^2 in this case), and d is the distance fallen (hf in this case). Substituting the values into the equation, we have: (61 m/s)^2 = (0 m/s)^2 + 2(-3.7 m/s^2)(hf). Solving for hf, we get: hf = (61 m/s)^2 / (2(-3.7 m/s^2)). Calculating this, we find that hf ≈ 240.56 m.

Answer 2

x=502.837m

v=61m/S

Step-by-step explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t (1)

{Vf^{2}-Vo^2}/{2.a} =X (2)

X=Xo+ VoT+0.5at^{2} (3)

X=(Vf+Vo)T/2 (4)

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

for the part a we can use the ecuation number 2

a=3.7m/s^2

Vf=61m/s

Vo=0m/s

`(Vf^(2)-Vo^2)/(2.a) =X \\`

`(61^(2)-0^2)/(2(3.7)) =X\\x=502.84m\\`

for the part b

the problem indicates that the speed with which it touches the ground is 61m / s