<h3><u>Answer</u>;</h3>
= F0 L ( 1 - 1/e )
<h3><u>Explanation;</u></h3>
Work done is given as the product of force and distance.
In this case;
Work done = ∫︎ F(x) dx
= F0 ∫︎ e^(-x/L) dx
= F0 [ -L e^(-x/L) ] between 0 and L
= F0 L ( 1 - 1/e )
Answer:
The change in momentum is
Explanation:
From the question we are told that
The mass of the probe is 
The location of the prob at time t = 22.9 s is 
The momentum at time t = 22.9 s is
The net force on the probe is 
Generally the change in momentum is mathematically represented as

The initial time is 22.6 s
The final time is 22.9 s
Substituting values

Answer:

Explanation:
<u>Vertical Launch Upwards</u>
In a vertical launch upwards, an object is launched vertically up from a height H without taking into consideration any kind of friction with the air.
If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

The object referred to in the question is thrown from a height H=0 and the maximum height is hm=77.5 m.
(a)
To find the initial speed we solve for vo:



(b)
The maximum time or the time taken by the object to reach its highest point is calculated as follows:


