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maw [93]
3 years ago
9

Consider a projectile launched with an initial velocity of v0 = 120 ft/s, inclined at an angle, θ with the horizontal. Let us as

sume that the projectile lands on a spot that is at a height h = 10 feet, raised in elevation from its launch point. For this entire problem, you can neglect drag.
a) Determine an expression for the horizontal range R as a function of time. Here, range represents the total horizontal distance traveled from launch point to point of landing. Your answer will be in terms of initial velocity v0, time t, height h and acceleration due to gravity, g. At this point, do not substitute any of the numerical values yet.
b) Determine the time taken to achieve the maximum possible range, R. Think about maxima and minima from calculus when solving this part of the problem.
c) Once you have determined the time from part (b), substitute the numerical values for the terms and obtain the actual time in seconds. Then, determine the initial launch angle, θ (also called the angle of attack), such that the projectile achieves the maximum possible horizontal range.
d) What is the numerical value of this maximum R (based on your calculations from parts (b) and (d)).
e) Determine the velocity at the peak of the trajectory of the projectile.
f) Determine the time taken for the projectile to reach its peak height on its trajectory.
g) Determine the peak height reached by the projectile.
Physics
1 answer:
Natali [406]3 years ago
4 0

Answer:

How to find the maximum height of a projectile?

if α = 90°, then the formula simplifies to: hmax = h + V₀² / (2 * g) and the time of flight is the longest. ...

if α = 45°, then the equation may be written as: ...

if α = 0°, then vertical velocity is equal to 0 (Vy = 0), and that's the case of horizontal projectile motion.

Explanation:

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Consider identical spherical conducting space ships in deep space where gravitational fields from other bodies are negligible co
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Answer:

 q = 8.61 10⁻¹¹ m

charge does not depend on the distance between the two ships.

it is a very small charge value so it should be easy to create in each one

Explanation:

In this exercise we have two forces in balance: the electric force and the gravitational force

          F_e -F_g = 0

          F_e = F_g

Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.

Let's write Coulomb's law and gravitational attraction

         k \frac{q_1q_2}{r^2} = G \frac{m_1m_2}{r^2}

In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.

          k q² = G m²

          q = \sqrt{ \frac{G}{k} }    m

we substitute

           q = \sqrt{ \frac{ 6.67 \ 10^{-11}}{8.99 \ 10^{9}} }   m

            q = \sqrt{0.7419 \ 10^{-20}}   m

           q = 0.861 10⁻¹⁰ m

           q = 8.61 10⁻¹¹ m

This amount of charge does not depend on the distance between the two ships.

It is also proportional to the mass of the ships with the proportionality factor found.

Suppose the ships have a mass of m = 1000 kg, let's find the cargo

            q = 8.61 10⁻¹¹ 10³

            q = 8.61 10⁻⁸ C

             

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A 60-kg box is being pushed a distance of 7.9 m across the floor by a force Upper POverscript right-arrow EndScripts ⁢ whose mag
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Answer:

Explanation:

Given

mass of box=60 kg

distance=7.9 m

\vec{P}=194 N

Force is parallel to displacement

coefficient of kinetic energy \mu =0.23

Work done by Force

W=P\cdot x=Px\cos \theta

W=Px\cos 0

W=194\times 7.9=1532.6 N

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