What Is the Speed and Distance of a Ball Dropped from a Descending Helicopter?

[slappy]

Homework Statement

A ball is dropped from a helicopter that is descending steadily at 1.52 m/s.
(a) After 3.00s, what is the speed of the ball?
(b) How far is it below the helicopter?

Homework Equations

unsure

The Attempt at a Solution

don't know where to start

 

[rock.freak667]

Homework Equations

unsure

What equations do you know that involve displacement, velocity and time when acceleration is constant?

 

[slappy]

What equations do you know that involve displacement, velocity and time when acceleration is constant?

From my notes
X - X(0) = V(0)t + at^2/2

 

[rock.freak667]

From my notes
X - X(0) = V(0)t + at^2/2

Do your notes contain any others? Specifically that only relate velocity, acceleration and time?

 

[rwisz]

I would first identify the key variables and equations needed to solve this problem. In this case, the key variables are time, velocity, and distance. The equation that relates these variables is the equation for constant acceleration, which is given by d = v0t + 1/2at^2, where d is the distance, v0 is the initial velocity, a is the acceleration, and t is the time.

For part (a), we can use this equation to calculate the velocity of the ball after 3.00 seconds. We know that the initial velocity, v0, is equal to the velocity of the helicopter, which is descending steadily at 1.52 m/s. We also know that the acceleration, a, is equal to the acceleration due to gravity, which is approximately 9.8 m/s^2. Plugging these values into the equation, we get:

d = (1.52 m/s)(3.00 s) + 1/2(9.8 m/s^2)(3.00 s)^2
d = 4.56 m + 44.1 m
d = 48.66 m

Therefore, after 3.00 seconds, the ball will have fallen a distance of 48.66 meters. To find the velocity, we can use the equation v = v0 + at, where v is the final velocity. Plugging in the values, we get:

v = (1.52 m/s) + (9.8 m/s^2)(3.00 s)
v = 1.52 m/s + 29.4 m/s
v = 30.92 m/s

Therefore, after 3.00 seconds, the ball will have a velocity of 30.92 m/s.

For part (b), we can use the same equation, d = v0t + 1/2at^2, to calculate the distance the ball is below the helicopter. Since we already know the final distance, we can rearrange the equation to solve for the initial velocity, v0. We get:

v0 = (d - 1/2at^2)/t
v0 = (48.66 m - 1/2(9.8 m/s^2)(3.00 s)^2)/(3.00 s)
v0 = (48.66 m - 44.1 m)/