..................,,,,,,,,,
2 answers:
You might be interested in
Let us assume that rocket only runs in initial energy and not using its own to flying.
Also , let upward direction is +ve and downward direction is -ve .
Initial velocity , u = 58.8 m/s .
Acceleration due to gravity , .
Final velocity , v - = 0 m/s .
We know , by equation of motion .
Hence, this is the required solution .
You need to first measure the angle of descent, i.e. the angle the hill makes with the ground. Then identify the forces acting on the sled, split them up into horizontal and vertical components, or into components that are parallel and perpendicular to the hill, and use Newton's second law to determine the components of the sled's acceleration vector.
There are at least 2 forces acting on the sled:
• its weight, pointing downward with magnitude <em>W</em> = <em>m g</em>
• the normal force, pointing perpendicular to the hill and away from the ground with mag. <em>N</em>
The question doesn't specify, but there might also be friction to consider, indicated in the attachment by the vector <em>F</em> pointing parallel to the slope of the hill and opposing the direction of the sled's motion with mag. <em>F</em>.
Splitting up the forces into parallel/perpendicular components is less work. By Newton's second law, the net force (denoted with ∑ or "sigma" here) in a particular direction is equal to the mass of the sled times its acceleration in that direction:
∑ (//) = <em>W</em> (//) = <em>m</em> <em>a</em> (//)
∑ (⟂) = <em>W</em> (⟂) + <em>N</em> = <em>m </em><em>a</em> (⟂)
where, for instance, <em>W</em> (//) denotes the component of the sled's weight in the direction parallel to the hill, while <em>a</em> (⟂) denotes the component of the sled's acceleration perpendicular to the hill. If there is friction, you need to add -<em>F</em> to the first equation.
If the hill makes an angle of <em>θ</em> with flat ground, then <em>W</em> makes the same angle with the hill so that
<em>W</em> (//) = -<em>m g </em>sin(<em>θ</em>)
<em>W</em> (⟂) = -<em>m g</em> cos(<em>θ</em>)
So we have
<em>-m g </em>sin(<em>θ</em>) = <em>m</em> <em>a</em> (//) → <em>a</em> (//) = -<em>g </em>sin(<em>θ</em>)
<em>-m g</em> cos(<em>θ</em>) + <em>N</em> = <em>m </em><em>a</em> (⟂) → <em>a</em> (⟂) = 0
where the last equality follows from the fact that the normal force exactly opposes the perpendicular component of the weight. This is because the sled is moving along the slope of the hill, and not into the air or into the ground.
Then the acceleration vector is
<em>a</em> = <em>a</em> (//)
with magnitude
||<em>a</em>|| = <em>a</em> = <em>g </em>sin(<em>θ</em>).
