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

Answer:

x=502.837m

v=61m/S

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

$\frac{Vf^{2}-Vo^2}{2.a} =X$

$\frac{61^{2}-0^2}{2(3.7)} =X\\x=502.84m$

for the part b

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