A tiger leaps horizontally from a 7.5 m high rock with a speed of 4.5 m/s. How far will she land?
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.
759984
The total time for the rock to drop and the sound to come back is 2.40s.  The distance that the rock drops is (.5)(9.8 m/s^2)(time for rock to fall).  The distance that the sound comes back is (336 m/s)(time of sound to come back).  The two times added together are 2.4.\nThese two distances are the same so.\n(.5)(9.8m/s^2)(time for rock to fall)^2= (336 m/s)(2.4 s - time for rock to fall) \n(4.9m/s^2)(time)^2 = 806.4m - (336 m/s)(time for rock to fall)\n(4.9m/s^2)(time)^2 +336 m/s(time) -806.4 m = 0\nUsing the quadratic formula, the allowed values for time is\n2.31 s\nDistance down to water is (.5)(9.8 m/s^2)(2.31s)^2 = 26.15 m\n\nb) if you were to ignore travel time for the sound, you would simply have (.5)(9.8m/s^2)(2.4s)^2 = 28.22 m.  Your percentage error would be (28.22-26.15)/26.15 = 7.9%
473407
An object in motion remains in motion unless it is acted upon by an outside force.  Say you're traveling in a train going 200 miles an hour, jumping on a trampoline.  Before you jump, both you and the train are moving 200 miles an hour forward.  When you jump, you'll begin moving upward at about 5 mph, but you'll still be going 200 mph forward as well, just like the train.  When you come back down, you'll meet the trampoline exactly where you left, because you, the train, and the trampoline have all been moving forward at the exact same speed.  Here's another example:  The circumference of the earth is around 24,000 miles, and the earth completes one revolution per day.  That means that, in one day, you travel in a 24,000 mile circle, yet you don't feel that you are moving because everything else is moving at the exact same speed.
484307
in air= 340m/s\nin water=1500m/s\nin solid=4500m/s
397772
It was a female measuring approximately  feet or 7 meters.
626855
This is your garden variety related rates problem.\n\nAt any instant when the monkey is on the wall, you have yourself a right triangle, with the taut rope being the hypotenuse, the distance from fastening point to the wall being the horizontal leg, and the distance the monkey climbs vertically being the vertical leg. The one equation which relates all three of those quantities together is the Pythagorean Theorem: x^2 + y^2 = z^2 (x and y being the legs and z being the hypotenuse).\n\nSince this is a related rates problem, you need to differentiate this equation with respect to time (t). The result is 2x(dx/dt) + 2y(dy/dt) = 2z(dz/dt). All of the d*/dt terms represent the rate * is changing over time.\n\nSo, now it's a matter of plugging in the appropriate values given in the word problem. According to the word problem, x=8 ft, dy/dt=4 ft/s, y=6 ft (at that instant). From this info, one can derive that z=10 ft (via Pythagorean Theorem) and dx/dt=0 ft/s (because the distance between the fasten point and the wall is fixed). The problem is asking for dz/dt. Now, plug in:\n2(8)(0) + 2(6)(4) = 2(10)(dz/dt)\nSolving for dz/dt, your answer should be 2.4 ft/s (if the math is correct).
316157
The correct formula to use is one of the simple equations of uniform linear motion.\n\nv² = u² + 2as\n\nu = initial velocity = 0\na = acceleration = 1.7 m/s²\ns = height = 500\n\nWhen rearranged\n\nv = sqrt(2as) = 41.2 m/s on the moon\n\nThe apple on earth will approach terminal velocity asymptotically due to air resistance. To prove the apple reached a faster velocity from this height on the moon is difficult without doing calculations involving air resistance. For a height higher than 500m it is trivial. The apple on the moon will reach a higher velocity before impact, whereas the apple on earth will still be limited to falling at its terminal velocity of 42m/s.
894321
The horizontal component of speed\n= 40 m/s * cos(30)\n=40 cos(30) m/s\nmass\n=0.15 kg\nSince at the highest point, vertical velocity=0, and horizontal velocity is unchanged (ignore air resistance), so \nK.E.=mv^2/2=0.15*(40cos(30))^2/2\n=0.15*1600*3/4/2\n=90 joules
147124
A cat is/is not in the box\nparticles wave goodbye\nI act and the reaction moves\nfrom the rainforest to the moon\nI drop at 72 feet per second squared\nand force my mass to pressure you.
948016
The range of the marble as it hits the floor, should be 5.5m.\nHere's how I figured it out :\nFirst, you would use the Dy = ViT + 1/2 gT^2 formula, to find your time. you should end up with 1.5 = 4.9T^2\nThen, divide the 1.5 by the 4.9, and take the square root of that number. You should end up with something like 0.55328.....\nSo, your time = 0.5s.\n\nThen, use the Dx = VxT formula to find your distance when it hits the floor.\nSo, Dx = 10m/sx0.5s\nand, Dx should equal 5.5m\nTherefore, it took the marble 0.5s traveling at a velocity of 10m/s to hit the floor 5.5m away after being launched horizontally off a table that's 1.5m high.\n\nHope this helps!
890149
that all depends... if u r building a dirt jump then u need to build it by feel. if the jump is 1.3 meters high and goes up at around a 45 degree angle and assume u r going full speed on a dirt jumping mtb then the jump will carry you up into the air at about 1.4 meters hight. and u will land about 2 meters away. but if u r building a wooden jump then you will be able to go higher and furthur but for mtbs dirt jumps work better because they are very heavy and would be harder to controll on a wooden jump than on dirt.
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!
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.
886536
If only birds really could fly in vacuums, but this is physics class world, not real world, so...\n\nIn the x direction the fish has a velocity of 18 m/s\nThis won't change according to Newton's laws.\n\nWe are interested in knowing how fast the fish is going when it hits the ground....I assume the bird dropped it.\n\nWhen the bird is carrying it it has 0 velocity in the y direction.\n\nWe know the height it falls from and the acceleration so we can use the equation below to find the final velocity in the y direction\n\nv = sqrt[V^2 + 2hg]  where V is the initial velocity in the y direction so\n\nv = sqrt[2*5.4*9.81]\nv = 10.293 m/s\n\nWe use the pythagorean theorem to add the vectors \n\nspeed = sqrt[18^2 + 10.293^2]\nspeed = 20.735 m/s
695472
Okay,\n\nQuick method:\n\nShe picks up a certain K.E. (kinetic energy) before she hits the water and loses this K.E. 3m under. \n\nWe have KE = Force x distance = mad\n\nso we have m x a1 x d1 = m x a2 x d2\n\nm =65kg, a1= 9.8m/sec, d1=13m, d2= 3m\n\nF(ave)= m x a2= m x a1 x d1/d2 = 2760 1/3 N \n\nThe force exerted on her by the water from when she hits until she stops 3m under will not be constant but the average force stops her in a distance of 3m and has the above value
782512
You need to use kinematics and Newton's 2nd law to solve the problem.  In the horizontal direction, Newton's 2nd law states:\n\nForce - friction = m*a.  Since you are given the values, you can solve for the accleration:\n\n1149 - 909 = (2700)*a, solving for a, you get:\n\na = 0.089 m/s^2\n\n----------------\n\nNow use kinematics to solve for the distance needed to reach the desired final velocity.  Use the following kinematic equation:\n\n2ax = v_f^2 - v_i^2.  You know that:\n\nv_f = 3.2 m/s\nv_i = 0 m/s (starting from rest)\na = 0.089 m/s^2\n\nNiw solve for x:\n\nx = [v_f^2 - v_i^2]/(2a) = 3.2^2/(2*0.089) = 57.6 meters\n\n---------------\n\nHope this helps
1168034
A tiger of course.\n\nYou could kick a foot long wasps ass....with a frying pan.\n\nA tiger would have you dead in seconds...