You shoot an arrow vertically into the air from 5 ft with an initial velocity of 96 ft/second...?
958247
v=v0-at=0=96-32t\n32t=96\nt=3sec\n\na=32'/s^2\nv0=96'/s\nd=5+96t-16t^2\nd=5+96*3-16*9=5+288-144=5+144=149'\n\nA) The arrow will reach a height of 149'\nit takes 3sec to go up & 3 more to come down to 5'\nthe last 5'\n5=96t+16t^2\n16t^2+96t-5=0\nt=(-96+/-√96^2+4*5*16)/32\nt=-3+/-√9216+320=-3+√9536/32\nt=-3+97.65/32=.05\nB) total time=3+3+.05=6.05sec
735664
we know m=0.0043kg, Vfinal=304m/s, d=0.86m and Vinitial=0\n\nwhat we need is the value of acceleration ( remember F=ma )\n\nso the formula we need is\n\nvfina^2=viinitial^2 + 2ad\n304^2=0 + 2*0.86*a\na=(304^2)/(2*0.86)\na=53730.23256\n\nso F=m*a=53730.23256*.0043=231.04 N\n\nForce is 231.04 N\n\n*****************************************************\n\nHere's an alterative if you want to double check  \n\nthe kinetic energy (KE)=1/2 *m*(velocity)^2\n\nthe KE of the bullet must equal the Work needed to fire it\n\nKE=0.5*0.0043*304^2 =198.6944 j\n\nWork=F*d=198.6944\n\nso F=(198.6944)/0.86=231.04 N\n\nAgain we get  force = 231.04 N\n\nSo we're pretty sure this is the correct answer
891399
assuming constant velocity, the dart is moving horizontally at a rate of 15m/sec at the same time it is accelerating downwards. \n\nso the horizontal distance = 15m/sec * t ot t=2/15 sec\n\nsince acceleration due to gravity is constant, \nvertical distance = (1/2)g(t^2)\n\nso the vertical distance it falls = .088 meters or 8.8 cm\n\nso it falls 8.8cm - 5 cm = 3.8 below the target
334471
If fired straight up, no the terminal velocity is too small.  But if fired at an angle, the bullet maintains horizontal speed and that can be enough to kill.
368641
I'm not sure what 500 above the horizontal means - radians maybe?  Or is there something missing, like a launch angle, and that you're 500 ft above ground when the ball is thrown?\n\nIf I'm picturing the scenario right, you're basically shooting a ball up at an angle and finding all of the specifics - max height, velocity at impact, etc.  All of those can be found using some of the following equations:\n\nvf^2 = vi^2 + 2ad\nvf = vi + at\nd = vt\n\nwhere\nvf = final velocity\nvi = initial velocity\na = acceleration due to gravity (9.8 m/s^2 or 32.2 ft/s^2)\nt = time\nd = distance\n\nHorizontal and vertical velocity are independent of each other.  The vertical velocity changes due to gravity, and the horizontal velocity will remain constant throughout the ball's flight (assuming no air resistance is in play).  So to find the max height, we'd need the vertical velocity component of the 110 ft/s.  That comes out to 110 sin x (where x is that launch angle I asked for earlier).  So this will be your vi in your first equation.  vf = 0 at the apex (the ball has stopped and is about to reverse direction), and since you are decelerating, use -32.2 ft/s^2 (I know you specified 9.8 m/s^2, but the units don't match the 110 ft/s - you can always convert one or the other if you like).  Solving that first equation will give you the distance it traveled.  If there is an additional height involved (remember if I asked if you were 500 ft above ground?) then your total height from ground level would be your answer from the equation plus 500.\n\nThe time to the apex can be found with the second equation.  Again, using vi = 110 sin x and vf = 0, you can solve for the time (and again, keep your units consistent or else the answer doesn't make sense).\n\nLet's jump to the last question about the velocity of the ball upon impact.  If it falls to the same height that it was thrown at, then the velocity on impact will be the same as the launch velocity - 110 ft/s.  But if there is that additional 500 ft height, we can use that first equation to find the vertical component of the impact velocity.  If we were to picture that part of the equation as starting from the apex, vi = 0.  The distance d will be the distance found in part (i) plus 500 ft.  So you can solve for the final velocity there - but remember that this only solves for that vertical component of the total velocity.  You'd need to find that total via the Pythagorean theorem - picture a triangle with the vertical velocity as one leg, and 100 cos x (the horizontal component of the original launch velocity) as the other leg.  Square both, add, and take the square root to find the total velocity on impact.\n\nOn to part (iii) - now that you have the final velocity (just use the vertical component as calculated above), you can use the second equation to find the time it took to fall from the apex to the ground.  Add this to the answer in part (ii) and you have the total time in flight.\n\nDown range, I'm assuming, is the horizontal distance the ball traveled.  Using the third equation, you can find that.  The velocity in this case is the horizontal component of the launch velocity - 100 cos x.  Take this times the time from part (iii) and you have the range.\n\nHope this helps - figure out what that launch angle is and how that 500 plays in to the equation and you should be on your way!
888567
a) zero momentum, as velocity on top is zero. p=mv\n     p=0.100kg * Om/s\n\nb) Vf = Vi + at\n      0=15+(-10)t\n     t=1.5ec (to reach maximum height)\n\nDivide by 2 for halfway up .i.e 0.75ec\n\nVf = Vi + at\nO=15 + (-10)(0.75)\nVf = 7.5m/s\n\np=mv   (p = momemtum)\n\np = (0.100kg)(7.5m/s)\np = 0.75kg*m/s, upwards (Direction is important)
317623
i believe the formula is\n\n16 foot per second per second\nor\n16ft + 32ft + 48ft + 64ft + ect....\n1 sec, 2 sec, 3 sec, 4 sec, ect....\nadding 16 foot for each additional second\nuntil you are adding the terminal velocity for each second.\nthe terminal velocity is the fastest something can fall.\nthe terminal velocity changes with the shape and mass\nof the falling object.\nexample : a piece of paper vs a sphere of lead\nyour ball would be approx. 96 ft of the way down\nor 4 ft off the ground, provided it wasn't Styrofoam.
799812
The maximum height is given by the point where the slope (derivative) is zero.\n\nh'(t) = -32t + 96 = 0\n32t= 96\nt = 3\n\nSo the maximum point will be at time t = 3.\n\nh(3) = 5 + 96(3) - 16(3)²\nh(3) = 5 + 288 - 16*9\nh(3) = 293 - 144\nh(3) = 149 ft.\n\nNow just don't stand directly underneath this when the arrow returns to Earth!
978760
Projectile Motion-  Examine the initial velocity and angle necessary to hit targets at different points.
208645
All objects accelerate downward at a rate of 9.8 m/s/s.  If the object falls 2.6 m, at that rate, it would it the ground in approximately .53 seconds at a speed of 5.2m/s. To calculate it falls approx. 4.9m the first second then continues to speed up at a rate of 9.8m/s/s after that.  Since 4.9m is farther than the distance fallen, divide 2.6 by 4.9 and get the time taken to fall.  Take that answer and multiply that by the constant rate of acceleration to get 5.2m/s
960647
Okay so if you are falling towards the ground the problem is that you are going to have a lot of speed when you hit the ground and that's not a good thing at all.  It is true that you can use your legs to push yourself upwards.  If you could push yourself upwards enough you could reduce your velocity towards the ground or even cancel it out!  So the question is how much upwards velocity can you give yourself by jumping.\n\nSo a little background first.  Say a slingshot fires a ball at 50 m/s.  If you were falling straight down towards the ground at 40 m/s and you shot the ball straight up it would have a velocity of 10 m/s upwards.  How did I get that? I got it because you can just add and subtract velocities like that if they are in the exact same direction (or exactly opposite direction).  I walk left at 5m/s along the ground and throw a ball same way at 6m/s it is then going left at 5ms+6m/s.  Or if I threw it right at 6m/s while walking left it would start with a velocity of (5m/s-6m/s) left which is -1m/s left (that's the exact same thing as 1m/s right).\n\nYou can figure out how much upwards velocity you can give yourself by jumping using this formula:\n\nvf²=vi²+2ad\n\nvf = final velocity, vi = initial velocity, a=acceleration, d=distance\n\nImagine you're standing on the ground while we think about this for a second and that you are great at jumping and you can jump 1 meter upwards.  You start with some velocity vi from your jump and have 0 velocity when you hit the peak of your jump at 1 meter in the air.\n\n0=vi²+2(9.8)(1)\n\nvi=4.4m/s\n\nSo if you are a strong jumper you can give yourself an upwards velocity of 4.4m/s.\n\nNow finally back to the elevator, if you and the elevator fell from a 5th floor at about 15m height you would be going down pretty quickly, almost 20m/s before you hit the ground.\n\nNow if you could jump upwards JUST before hitting the ground you might at best get almost 4.4m/s of upwards velocity and you could subtract that from your downards velocity of 20m/s.  That gives about 15m/s and that is a big problem to hit the ground at that speed.\n\nUnfortunately you wouldn't even do that well because not all of the force you applied to jump upwards would go into accelerating you upwards.  Some of it would go into pushing the elevator downwards even faster so you would gain even less upwards velocity.\n\nThe short answer is you could reduce your speed a little bit but not nearly enough to stop yourself from landing way too hard.
141479
You should do your own homework problems.  Being told answers will not help you learn.\n\nI'll try to help you understand, though.\n\nThink about this.  The tiger's speed accelerates at a rate of 9.81 m/s^2.  For constant acceleration, distance is related to time by the equation d=t*v_0+(t^2)*a where v_0 is the initial speed IN THE DIRECTION OF ACCELERATION, d is the distance traveled, a is the rate of acceleration and t is the time.  Use that to figure out how much time it takes the tiger to reach the ground.  Then multiply this amount of time by the forward speed.  This tells you how far the tiger has moved forward upon landing.\n\nFor the ball the same formula applies.  It works for the projectile too, but you must decompose the velocity into its vertical and horizontal components (hint:  one will be the sine of the angle it's fired from times the speed and the other will be the cosine of that angle times the speed).  I'm going to leave it up to you to figure out which.  This question obviously has more to do with wanting your homework answered for you than with personal curiosity, because of its specificity, and if you just didn't get the concepts it's ok.  It is not ok to submit your homework on the internet in hopes that others will do it for you.
695797
Assuming that 0.5% means 0.5% of the way straight down, or 0.5% of 90 degrees = 0.45 degrees, and that gravity is neglected due to its insignificance over that short a time, and that 90 mph = 132 fps,\n\nx = 132t cos 0.45\ny = 5 -132t sin 0.45\n\nwhen x = 56, t is approximately 0.425 sec and y is approx. 4.56 ft
821062
I think you are supposed to assume the minimum velocity possible, which means tossed at a 45 degree angle. Otherwise he could throw it farther by using the same velocity at 45 degrees.\n\nvelocity = v, y velocity at time t = v/sqrt(2) - 10t so y velocity is 0 at \n\nt = v/(10sqrt(2))\nIn time 2t the distance travelled is 96.90, so\n\n96.90 = 2tv/sqrt(2)\n\nThese are 2 equations with 2 unknowns, solve for t and v (there may be a careless mistake in the details since I did this fast, but the technique will work)
730340
Projectile range would be the equivalent of horizontal distance.\nThe forces involved are gravity (acting on the projectile's mass in the vertical direction), and the "push" from the initial thrust.\n\nAs the angle of the firing position increases above the horizontal, more of the initial push is used to combat gravity.\n\nIf the angle is too low, the projectile will meet the ground before it's maximum range (not enough action against gravity).\n\nFor most projectiles, the optimum angle is between 42 and 45 degrees (depending on the strength of the initial impetus). This is because the trajectory approximates a parabola.\n\nWith the given choices, the answer will have to be 45 degrees (answer B)...
516788
Rollo's bullet hits the ground first, with or without taking air resistance into account. \nThe dropped bullet starts at zero speed and increases it's speed as it falls. The fired bullet starts at high speed and (neglecting air resistance) increases it's speed even more as it falls. Since the fired bullet starts out faster and keeps going faster than the dropped bullet all the way to the ground, it gets there first.\n\nNow let's include air resistance. The fired bullet will still start at high speed but it will start slowing down right away due to friction with the air. At some point it would have slowed to terminal velocity, the velocity at which air resistance equals the force of gravity, and then it will then fall the rest of the way at that speed. The dropped bullet will start from zero speed and speed up until it reaches terminal velocity and will then fall that that speed for the rest of the way. So, although both bullets will hit the ground at the same speed, the fired one traveled part of the way faster than the dropped bullet and never slower than the dropped bullet and still hits the ground first, although the difference is not as great as in the no air resistance case.\n\n(EDIT) Oh, I agree with dei48399. You probably meant he fired the bullet directly to the side. Nobody would doubt the outcome it it was fired down, would they?
759333
all object fall 4.9m after 1 sec\n\nX=1/2(9.8m/s^2)(2s)^2\nx=19.6m\n\nx=(1/2)(9.8m/s^2)(3s)^2\nx=44.1m