Answer:
=9.72 m/s
Explanation:
From the Newton's laws of motion;
x=2(v²cos∅sin∅)/g
Using geometry we see that 2 cos∅sin∅ = sin 2∅
Therefore, x= (v²sin 2∅)g, where v is the take off speed x the range and ∅ the launch angle.
Making v the subject of the formula we obtain the following equation.
v=√{xg /(sin 2∅)}
x=7.80
∅=27.0
v=√{7.8×9.8/sin(27×2)}
v=√94.485
v=9.72 m/s
Answer:
1st law--Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. This is normally taken as the definition of inertia. The key point here is that if there is no net forceacting on an object (if all the external forces cancel each other out) then the object will maintain a constant velocity. If that velocity is zero, then the object remains at rest. If an external force is applied, the velocity will change because of the force.
2nd— The second law explains how the velocity of an object changes when it is subjected to an external force. The law defines a force to be equal to change in momentum(mass times velocity) per change in time. Newton also developed the calculus of mathematics, and the "changes" expressed in the second law are most accurately defined in differential forms. (Calculus can also be used to determine the velocity and location variations experienced by an object subjected to an external force.) For an object with a constant mass m, the second law states that the force F is the product of an object's mass and its acceleration a:
F = m * a
For an external applied force, the change in velocity depends on the mass of the object. A force will cause a change in velocity; and likewise, a change in velocity will generate a force. The equation works both ways.
3rd law-- The third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal force on object A. Notice that the forces are exerted on different objects. The third law can be used to explain the generation of lift by a wing and the production of thrust by a jet engine.
a) Cumulus is 100% the correct answer
Answer:
The question is missing something it doesn't say how fast down its going and doesn't show the figure sorry for wasting an answer
