What was the speed of a brick when it hit the ground if dropped from 400 feet.?
374463
Assuming no air resistance, the first equation is:\n\n400=1/2 a t^2  (a, as you said, is 32.2 ft/sec^2)\n\nt, the time calculates to be abot 4.98 seconds\n\nthen, v=at, gives v=(32.2)*(4.98)=160 ft/sec\n\nIn the real world, the speed after 400 feet of fall will be less than that due to air resistance, which would slow a brick considerably at 160 ft/sec (over 100 miles/hr)
767639
Don't things fall at 32ft/sec.
955270
I won't do it for you, but here is what you need to do:\n\nFind the height at which the explosion occurs = h.\n\nFind the horizontal component of momentum at the instant before the explosion (which is actually the entire momentum, since there is no velocity in the vertical direction at that instant!) = Mtot x Vhoriz =m1 (Mtot is the entire shell).......(1)\n\nNow, the total momentum the instant after the explosion is:\n\n(M1 x 0) + (M2 x V2) .......(2)(since the large part of the shell stops travelling horizontally, and is already not travelling vertically) Where V2 is the horizontal velocity of the small part of the shell.\n\nSince momentum is conserved, these two quantities are equal, so:\n\nMtot x Vhoriz = M2 x V2, so you can calculate V2, the new horizontal velocity of the small part of the shell.\n\nNow, simply calculate the time taken for the shell to travel vertically to the ground using s = 1/2at^2, and from V2, you can calculate the horizontal distance travelleed by the small part of the shell...presto!
172926
The others gave the correct answer overall, but got the details wrong.  All objects accelerate, when dropped in free fall near the surface of the earth, at 9.8 meters/second/second (not 9.8 inches/s or 9.8 miles/hour).\nThe reason heavier object don't fall faster is that, although they have more weight, i.e. more gravity pulling on them, they also have more mass, i.e. more "inertia", which is the resistance of objects to being accelerated.  The two things exactly cancel out; isn't that convenient?  \n\nBtw, the reason why they cancel out, that is, the reason why the mass that shows up in the formulas for gravity is the same quantity that shows up as inertia in Newton's laws, was not fully solved until Einstein, but the fact that they do was known to Galileo.
1119837
First divide up the initial velocity into its horizontal and vertical components.\nV is the initial velocity of 40 m/s.\nTheta is the launch angle of 37 degrees above the horizon.\nV_x_i and V_y_i are the horizontal and vertical components of the initial velocity respectively.\n\nV_x_i = V * cos(theta) = 40 * cos (37) = 31.95 m/s\nV_y_i = V * sin(theta) = 40 * sin (37) = 24.07 m/s\n\nAlthough the vertical component of the object's velocity will change as the object experiences gravitational acceleration toward the ground, neglecting air resistance, the horizontal component of velocity can be treated as a constant.\n\nV_x = V_x_i = 31.95 m/s\nV_y = V_y_i  - g*t\nwhere t is the time after the object is launched and g is the gravitational acceleration experienced by the object (g = 9.81 m/s^2).\n\nIf the fence which the object hits is 128 meters away (in the +X direction), then the object (traveling at 31.95 m/s in the +X direction) will hit it at about 4.01 seconds after launch.\n\nBy taking this value (4.01 s) and plugging it in for t in the below shown equation to solve for the distance relative to the initial launch height the object has traveled,\nd = V_y_i * t + 1/2 (g)t^2\none can calculate that the object is at a height of about 17.65 meters above its original launch height.\n\nSo I calculate the impact height to be 17.65 meters.
77860
hey, didnt i answer something very similar to this for you? you need to do F=ma so 43000000N/2100000kg then you get an acceleration, but then you have to subtract gravity from this because thats the acceleration produced by the rocket but there is still gravity pulling with 9.8 m/s^2 in the opposite direction. so just do (43000000/2100000) - 9.8, and that's your answer :)
1164981
I think you have left off part of the question. The speed of a ball is a vector. When you hit the ball, the vector points upward and forward. Gravity pulls the vector downward. Small forces (air friction, etc.) retard the total speed, but assume those can be ignored.\n\nAt some point, the pull of gravity will pull the vector to point down instead of up. That's the maximum height. The horizontal component of the speed vector should not change (given our assumptions). It still goes forward at the initial velocity. However, The vertical component is instantaenously at zero. In a moment, the ball will start to accelerate vertically and the vector will continue to pull down below the horizontal line becuase of the force of gravity.\n\nIf this were not true, the ball would reach maximum height and then just drop from the sky (0 overall velocity) which you know it does not do.\n\nSo of your choices, probably equal to initial or ever so slightly les than initial would be the best.\n\nThe same thing happens if you drop, say, a ball from 10 feet and fire a bullet from a height of 10 feet. The both hit the ground at the same time.The fact that the bullet is speeding horizontally doesn't affect gravity pulling on it vertically. (yeah small effects like curvature of the Earth, wind resistance, etc. being ignored).
123981
First find the new velocity of the clay/block combo using conservation of momentum (Vc)(Mc) = (Mb + Mc)V  , solve for V.  Impulse = (M)(change in velocity).  Your mass is .02, and the change in velocity is (30 - V).  This gives you an impulse of .593 Ns.
927051
The given final speed doesn't seem to compute. I get 14.8 m/s. \nMa*Va + Mb*Vb = (Ma+Mb)*Vf \nis the formula that applies to show conservation of momentum in a completely inelastic collision, which is what you have. The only explanation I could suspect is some cargo was spilled. \n\nTo continue the assignment - kinetic energy is given by\n(1/2)*m*v^2\nThere is no requirement that all the original kinetic energy be measurable as kinetic energy after a completely inelastic collision. But you should be able to point to where the missing energy went. Some of the energy probably went into cart damage. Perhaps the cargo slid around.
837767
Bingo!!! You got it...air resistance.  If you dropped them in a total vacuum, they would both fall at the same rate.
1118371
double_nubbins has the right approach.  I will add a couple of hints: how long will it take the car to fall 19ft (remember the acceleration equation s=.5*g*t^2).  Assuming there is no air drag, the horizontal velocity of the car must be enough to cover the 52m width in that time.\n\nSpeed of the car? The horizontal speed? The falling speed?  The speed in the direction of motion?\n\nHorizontal speed: as calculated above\nFalling speed = g*t, where t is the time calculated above.\nSpeed  in the direction of motion = sqrt(horiz speed ^2 = falling speed ^2)
487670
None. It takes no power to maintain that speed (on level ground, no friction). The object has kinetic energy (0.5 * m * v^2). Power was required to put it in motion and power can be recovered by decelerating it. If you do the math, the energy is 3551.86 ft-lb, equivalent to 1 horsepower for 6.458 seconds. Answer 1 neglected the fact that weight must be converted to mass by dividing by g, and energy is proportional to v^2.
824568
We can use trigonometry to determine the vertical and horizontal components of the stone's velocity.  This will let us determine how far it goes.\n\nImagine a right triangle with a hypotenuse of 14 units and one angle of 26 degrees.  The side opposite the 26 degree angle will be as long as the stone's vertical velocity.  The side adjacent to the 26 degree angle will be as long as the stone's horizontal velocity.\n\nSince:\n\nsinA = opposite / hypotenuse... \nsin26 = vertical velocity / 14...\nvertical velocity = 6.14m/s\n\ncosA = adjacent / hypotenuse;\ncos26 = horizontal velocity / 14...\nhorizontal velocity = 12.56m/s\n\nIf the object had an initial vertical velocity of 6.14m/s, we know how long it stayed in the air because:\nacceleration = change in velocity / time\nWhich is the same as saying:\ntime = change in velocity / acceleration\nThis applies in any situation where the acceleration is constant.\nSo, if we want to know how long it took for the stone to reach 0m/s vertical velocity (the point at which it stops rising and starts falling), the change in velocity will be 6.14m/s.  If we just divide this number by acceleration due to Earth's gravity (9.81m/s^2), we get 0.63 seconds.  The stone traveled upward for 0.63 seconds.  This means that it must have also traveled downward for 0.63 seconds before hitting the ground, for a total flight time of 1.26 seconds.\n\nNow, we have a simple velocity equation.  Since we know that:\ndistance = velocity * time \nSince we're only worried about how far the stone traveled horizontally, we use the stone's horizontal velocity: 12.56m/s.  Time will be the total time that the stone was moving: 1.26sec.\n\nSo...\ndistance = 12.56 * 1.26 = 15.83 meters.\n\nHope that helped!
19024
I'm explaining this only on condition you don't try the experiment yourself.\n\nGiven that the Empire State Building is 1,250 feet tall and ignoring such factors as wind resistance for the moment, a penny dropped from the top would hit the ground in approximately 8.8 seconds, having reach a speed of roughly 280 feet per second.\n\nThis is not particularly fast. A low-powered .22 or .25 caliber handgun bullet, to which a penny is vaguely comparable in terms of mass, typically has a muzzle velocity of 800 to 1,100 FPS, with maybe 75 foot-pounds of energy.\n\nOn top of this we must consider that the penny would probably tumble while falling, and that the Empire State Building, like all tall buildings, is surrounded by strong updrafts. As a result the penny's descent would be substantially slowed.\n\nThus while you might conceivably inflict a fractured skull on some hapless New Yorker (or, more likely, some cretinous tourist from Towson), the penny would not "go through just like that." I bet it wouldn't even penetrate the skin. Not that I intend to find out.\n\nFor the record, the Empire State folks claim no one has ever dropped anything off their building. Yeah, right.\n-By Cecil Adams
309904
There is a speed at which a falling object can't fall any faster, because the wind resistance holds it back.  It's called "terminal velocity".  It's different for a rock and a wad of paper the same size.  The terminal velocity for a raindrop is pretty low.
896561
If the object starts out at a height h0 and is dropped, then all the potential energy is converted in kinetic energy\n\nm*g*h0 = 0.5*m*v0^2\n\nso that at impact the object is moving at a speed v0\n\nv0 = sqrt(2*g*h0)\n\nAfter bouncing off the surface the object is moving with a speed v1\n\nv1 = e*v0\n\nwhere e is the coefficient of restitution. In a purely elastic collision e=1, but in the real world e<1. Using the conservation of energy again\n\n0.5*m*v1^2 = m*g*h1\n\nwhere h1 is the new height to which the object bounces.\n\nh1 = 0.5*e^2*v0^2/g\n\nAt moment of next impact the object is moving at speed v1, but rebounds at speed v2\n\nv2 = e*v1 = e^2*v0\n\nThe height h2 is then\n\nh2 = 0.5*e^2*v1^2/g = 0.5*e^4*v0^2/g\n\nThis continues for however many bounces you wish. Recall now that v0^2 = 2*g*h0 so that the heights are\n\nh0 = e^0*h0 (shown only for consistency with the following)\nh1 = e^2*h0\nh2 = e^4*h0\nh3 = e^6*h0\n...\nhn = e^(2n)*h0
64850
A correction to the first answer. The formula terminal V = SQRT(2W/(Cd*R*A)) is correct, as it expresses the fact that W is balanced by the drag force V^2*Cd*R*A/2. However, for two same-sized smooth balls of different materials, only W changes; Cd stays the same. The air density R depends only on altitude.\nSince air density increases with decreasing altitude, the terminal speed decreases during a fall. If dropped from high enough, the actual fall speed, which initially = 0, increases until the (decreasing) terminal speed is reached, and then the sign of the net force (gravitational + drag) changes so the falling ball slows. For the remainder of the fall, the terminal and fall speeds both decrease, but the fall speed is slightly faster. This speed difference provides the net upward force to slow the ball.