Partial preview of the text
Download projectile motion and gravity free fall and more Study notes Physics in PDF only on Docsity!
Theortes About Gravity Theories About Gravity Aristotle ® explained that an object tends to move towards its “natural resting place” ® heavier objects fall faster and lighter objects fall slower Galileo Galilei ® one of the firsts to perform experiments in studying gravitation debunk Aristotle’s earlier hypothesis and concluded that objects fall at the same rate regardless of mass Isaac Newton ® established the Universal Law of Gravitation Albert Einstein established General Theory of Relativity where gravity is described as the curvature of space-time continuum due to presence of a massive object Universal Law of Gravitation ® any two objects in the universe exert gravitational forces to one another ® these forces are equal in magnitude and opposite in directions (Newton’s Third Law) ® the magnitude of these forces are directly proportional to the product of their masses and inversely proportional to the distance between them Gravitational Field (g) © a force field that surrounds an object that exerts gravitational force © uses the same variable as acceleration due to gravity (g) because they are always equal in magnitude © the unit of gravitational field strength is m/s? Force Field ® made of infinite number of vectors called field vectors that surround and are pointing towards the object aside from gravity, magnetic force and coulomb (electric) force also involve force fields A cube and a sphere, with equal mass of 4.0 kg separated by a 2.0 m distance, are shown. If the length of the cube is 1.0 m, and the diameter of the sphere is 1.2 m, what is the gravitational force between the two? A 1200.0 kg asteroid floats freely in outer space. What is the magnitude of gravitational field at distance of 2.0 m from the asteroid’s center of gravity? If a 700.0 kg asteroid happens to pass by at distance of 2.0 m from the asteroid’s center of gravity, what will be the magnitude of gravitational force? M = 1200.0 kg m= 700.0 kg r=2.0m M 2 . g=G— g =6.67x10 tn.M_, (1200.0 kg) r kg? (2.0 m)? g =2.0x10 ms? Fo=mg Fg= (700.0kg)(2.0x 10~* mis? ) Fg=14x10 °N Free-fall Motion a special case of uniformly accelerated motion wherein gravity is the only force acting on an object it has the following assumptions: the motion is along the vertical the velocity is either upward or downward the weight is the only force acting on the object the acceleration is always g = 9.8 m/s? downward (or g = 980 cm/s? downward or g = 32.2 ft/s’ downward) air resistance or drag is not experienced can assume the motion as free-fall if: the air resistance is significantly weaker than gravity the object is close to Earth's surface such that g is very close to 9.8 m/s? Free-fall Motion t three scenarios that may happen in a free falling object which depends on its initial velocity scenario no. 1| If the object is dropped, it will have zero initial velocity and accelerates at 9.8 m/s? as it falls scenario no. 2| If the object is thrown downward (i.e., strike the floor), it will have non-zere initial velocity and accelerates at 9.8 m/s? as it falls scenario no. 3} If the object is thrown upward, it will gradually slow down on its way up until its speed becomes zero (at rest). The object will then fall, accelerating at 9.8 m/s? and moves opposite to the direction of its initial velocity Free-fall Motion © the following equations are used in dealing with free-fall: d=vt+—1g? gt=Vr— Mi 2gd = v,? — v2 1 d=vi-—gt? 2 2d = (v, + v,)t where g = Vi- Vi b. g= gt=Vy— Vv; vit gt= vy 0 + (980 cm/s2)(5.000s) = v, v,= 4900 cm/s downward v, = 4.900 « 10% cm/s 1 d=vit , gf d = (0)(5.000 s) + + (280 cm/s?)(5.000 s? d= 12,250 cm below d = 1.225 x 104 cm belo 1 co. d=vi+ gi? wit og 1 3.136 «104 cm = (O)f + 3 (980 cms’) 3.136 «104 cm = (490 cm/s?)f? p2 _ 3.136 10% em 490 cm/s? P= 64.00 s? t=+8.000s 2gd =v? -v? 2(980 cm/s?)( 3.136 104 cm) = v2 - (0 Vv? = 61,465,600 cm/s? v, = 7,840 cmis A bullet was fired with initial velocity of directly upward. a. How high is the bullet from its starting point after it was fired? b. What is the velocity of the bullet after it was fired? c. What is its acceleration after it was fired? d. What is its velocity at the peak of its flight? e. What is its acceleration at the peak of its flight? f. What is its acceleration after it was fired? g. What is its velocity after it was fired? h. At what time(s) is it above the ground? v, = 966 ft/s upward = 966 ft/s g = 32.2 ft/s” downward = -32.2 fi/s® a f=10.0s 1 oe d=vit ae d = (966 ft/s)(10.0 s) + 3 (-82.2 T's?) (10.0 s)2 d= 9,660 ft + (-1,610 ft) d=8,050ft | d= 8,050 ft above the starting point | Vr— Vi b gee (966 ft/s) + (-32.2 ft/s?) (10.0 s) = v, gt= vp, vt gt= vy c. g = 32.2 fs? downward 9. g = 32.2 ft's” downward v= Vv, + gt @ Vr Vosan = 0 v,= 966 ft/s + (-32.2 fi/s?)(50.0's) Vpeak =0 v, = 966 ft/s + (-1,610 ft/s) @. g = 32.2 ft/s? downward v, = — 644 ft/s V,= 644 ft's downward fg = 32.2 ft/s? downward Aa Projectile Motion free-falling motion where the velocity has a horizontal component the only force acting on the object is gravity the path or trajectory of the motion is in the shape of parabola, or part of it initial velocity (vi) angle of elevation(q) maximum height (h) [highest altitude the object along its trajectory range (R) maximum horizontal displacement total time of flight (T) |total amount of time it takes for the projectile to travel Projectile Motio the range Ris maximum if 6 = 45° two angles of elevation 6, and 8s, will have the same range for the same v; if @, + 8) = 45° the horizontal component of velocity is constant. To solve for the horizontal components of projectile motion: d,, = vt Projectile Motio the vertical component of velocity is a case of free-fall; to solve for the vertical components of projectile motion: 1 jt = Vay — Vi, d, = vit + gt? g yo Ny 2ady = Vy 4 v,,t — —gt? v2 2d, = (Vy + sind An arrow leaves a bow at at angle of from the horizontal. What is its maximum range? Find the maximum height of the arrow? How long would it take the arrow to land on the other side? b. v= vy, sine c. T=2t, = (30.0 m/s) sin30° gt=Vy- Vy = (30.0 m/s) (0.5) v,,= 15.0 mis " t,-—e 2gd, = v,2 — vy? "9g 2(-9.8 m/s?)h = 0 - (15.0 mis)? (-19.6 m/s?)h = - 225 m/s? 225m*/s? 19.6m/s? 
