contestada

Problem 1: An object of mass m is projected horizontally from the top of a
building. The object's initial velocity is v, its initial height is H, and it travels
a horizontal distance D before striking the ground. Lett represent the time
that the object is a projectile.
(a) Write two equations that relate the quantities v, H, D, t, and fundamental
constants.
H
(b) Write a single equation that expresses D in terms of v, H, fundamental constants, and NOT time.
(c) Suppose the ball is launched a second time with greater initial velocity. What must the velocity be
multiplied by in order to double the horizontal landing distance D? Explain your reasoning.
(d) Suppose the ball is launched a third time with the initial velocity v, but from a greater initial height. What
must H be multiplied by in order to double the horizontal landing distance D? Explain your reasoning.
(e) In terms of the given quantities and fundamental constants, write an expression for the net force acting on
the object while it is a projectile. Explain your reasoning.
(f) In terms of the given quantities and fundamental constants, write an expression for the object's kinetic
energy just before striking the ground. Explain your reasoning.

Respuesta :

According to Newton's laws of motion and the principle of conservation of energy, using the variables v, H, D, and t we have;

[tex] (a)\: H = \frac{1}{2} \cdot g\cdot {t}^{2} [/tex]

[tex]v = \frac{D}{t} [/tex]

[tex] (b)\: D = v \times \sqrt{\frac{2\cdot H}{g}} [/tex]

(c) The velocity must be doubled

(d) The height, H, must be multiplied by 4

(e) F = m•g

(f) K.E. = m•g•H

Which laws can be used to find the equations?

Mass of the object = m

Initial (horizontal) velocity of the object = v

Initial height = H

Horizontal distance traveled = D

Time object is a projectile = t

(a) According to Newton's laws of motion, we have;

[tex]H = \frac{1}{2} \cdot g\cdot {t}^{2} [/tex]

[tex]v = \frac{D}{t} [/tex]

Where;

g = Acceleration due to gravity

(b) From the two equations above, we have;

[tex]t = \frac{D}{v} [/tex]

Which gives;

[tex]H = \frac{1}{2} \cdot g\cdot { \left( \frac{D}{v} \right ) }^{2} [/tex]

[tex] D = v \times \sqrt{\frac{2\cdot H}{g}} [/tex]

(c) From the equation in part b, we have that the horizontal distance, D, is directly proportional to the horizontal velocity, v, therefore;

  • The velocity must be doubled to 2•v to double the horizontal distance to 2•D

(d) Given that the velocity remains constant, and multiplying both sides of the equation in part b by 2 gives;

[tex] 2 \times D = 2 \times v \times \sqrt{\frac{2\cdot H}{g}} = v \times \sqrt{\frac{2\cdot 4 \times H}{g}} [/tex]

Therefore;

  • H must be multiplied by 4 in order to double the horizontal landing distance.

(e) The net force, F, acting on the object is the force of gravity, which is given by the equation;

  • F = m•g

(f) Just before the object strikes the ground, according to the law of conservation of energy, we have;

Kinetic energy, K.E. = Initial potential energy of the object at the top

Therefore;

  • K.E. = m•g•H

Learn more about Newton's laws of motion here:

https://brainly.com/question/10454047

#SPJ1

Otras preguntas