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/s^2.
a) Find the velocity with which the camera hits the ground.
b) Find the time required for it to hit the ground.
Ok so I'm not even sure of which equation to use for this problem. Maybe if someone just gave me a hint as to where to begin, I could get the rest...?
The post is 8 years old now, but for anyone viewing the page: The book is wrong because the unit for velocity is m/s, not s(seconds). Your answers are correct, Lindsay.
vfinal^2 = vinitial62 + 2 a * height
solve for vfinal
vfinal= vinitial + a*t
solve for time.
Memorize these equations.
That should read vinitial^2
my shift finger is getting weak.
Ok so I plugged in the numbers and got 42.1 for a) and -11.4 for b).
However, my book says that I got them switched...that 11.4 s is a) and -42.1 m/s is b). How did this happen?
To solve this problem, you can use the equations of motion for constant acceleration. Here are the steps to find the velocity and time of the camera hitting the ground:
a) To find the velocity with which the camera hits the ground, you can use the formula:
v = u + at
Where:
- v is the final velocity (which we need to find)
- u is the initial velocity (in this case, the camera is dropped, so the initial velocity is 0 m/s)
- a is the acceleration (given as -3.7 m/s^2)
- t is the time taken to hit the ground (which we need to find)
Since the camera is dropped, the initial velocity (u) is 0 m/s. Plugging in the values, the equation becomes:
v = 0 + (-3.7) x t
Simplifying:
v = -3.7t
This shows that the velocity at any time (t) is equal to -3.7 times the time (t).
b) To find the time required for the camera to hit the ground, you can start with the equation of motion:
s = ut + (1/2)at^2
Where:
- s is the displacement (in this case, the height of the cliff which is 239 m)
- u is the initial velocity (0 m/s)
- a is the acceleration (-3.7 m/s^2)
- t is the time taken to hit the ground (which we need to find)
Rearranging the equation, you get:
s = (1/2)at^2
Plugging in the values:
239 = (1/2) x (-3.7) x t^2
Simplifying:
239 = (-1.85) t^2
Now, solve for t:
t^2 = 239 / (-1.85)
t^2 ≈ -129.19
Since this is a physical problem, we know that time cannot be negative. Therefore, there is an error in the arithmetic calculation or earlier steps of the problem. Please review the values and calculations you used to determine if any mistakes were made.