a pitch thrown horizontally at 147.84ft/s how far would it fall vertically when it reached the plate 60ft away
957111
The angle is given. Horizontally means with an angle of 0 degrees elevation.\n\nYou can find how long the ball takes to travel 60 ft horizontally (60/147.84 seconds), so, knowing the acceleration due to gravity (which you will have been given), you can calculate how far it must fall in that time.\n\nt = 60/147.84 = 0.405844156\n\ndistance d covered by falling body in t seconds = ½gt²\n\ng = 32.174 ft/sec²\n\nd = 0.5 * 32.174 * 0.405844156 * 0.405844156 = 2.649681386 ft
256497
around 1/6th that of earth so around 1.64m/s
534824
60mph is 88 ft/sec; (memorize that).\nSo, 90mph is 132 ft/sec.\n\nIgnoring air friction (which DOES have an effect!), it takes 90/132 of a second for the ball to travel that far - or about .682 seconds.\n\nHow far does something drop in about 2/33 of a second? Several feet.
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!
418402
Using one of the equations of linear motion:\n\nv² = u² + 2as\n\nInitial speed u = 0\nAcceleration a = 32 ft/s\nDistance s = 9 ft\n\nSolve for final velocity v = sqrt( 2 * 32 * 9) = 6.2 ft/s\n\nThis assumes negligible air friction.
714803
The equation necessary to find the horizontal and vertical positions of the stone are\nm horizontal = (Initial horizontal velocity)(time) + (.5)(horizontal acceleration)(time)^2\nm vertical = (Initial vertical velocity)(time) + (.5)(vertical acceleration)(time)^2\nSince your horizontal acceleration is zero, that part goes away.\nYour vertical acceleration is caused by gravity, so it should be about (-9.8m/s^2).\nYour Initial horizontal velocity is (28.0 m/s)(cos (46.0)) = 19.45 m/s\nYour Initial vertical velocity is (28.0 m/s)(sin (46)) = 20.14 m/s\nThus, your two equations are\nm (horizontal) = (19.45 m/s) (time) and\nm (vertical) = (20.14 m/s)(time) + (.5)(-9.8m/s^2)(time)^2\nPlugging in your numbers, you get\na) m(horizontal) = 19.45 m; m(vertical) = 15.24 m\nb) m(horizontal) = 33.1 m; m(vertical) = 25.9 m\nc) m(horizontal) = 79.75 m; m(vertical) = 0 m (back on the ground)
970380
Let rock be thrown at an angle of θ at a velocity of Vo\n\nThen vertical component = Vo sinθ\nand horizontal component = Vo cosθ\n\nAlso acceleration in horizontal direction = 0 \nand acceleration in vertical direction = -g = -9.8 m/s²\n\nVertical direction\n\na = -9.8\nv = u + at\nSo Vt = Vosinθ - 9.8t\nand s = ut + ½at²\nSo h = Vosinθ*t - 4.9t²\n= 0 when it is at ground level\nie t(Vosinθ - 4.9t) = 0\nSo t = 0 or Vosinθ/4.9 ... ****\nSo it will take Vosinθ/4.9 s to cross the river.\n\nHorizontally\n\nVt = Vocosθ + 0 (as a = 0)\n= Vocosθ\n\ns = ut + ½at²\n= Vocosθ * t\n\nSo 75 = Vocosθ *Vosinθ/4.9\n= Vo².sinθcosθ / 4.9\n= Vo² * 2sinθcosθ / 2*4.9\n= Vo² * sin2θ/9.8 \nThus Vo² * sin2θ = 75*9.8 = 735\nThus Vo² = 735/sin2θ\nNow least value of Vo occurs when Vo² is minimum and this will occur when sin2θ is maximum (=1)\n\nSo Min possible Vo = √735 \n≈ 27.11 m/s (projected at 2θ = 90° ie θ = 45°)\n\nMinimum time = Vo*sin45°/4.9 (from ****)\n≈ 3.91 s
735384
Using the law of conservation of energy determin\n\na) the vertical component of vector V(initial)\n\nIt went up 20 meters, mgh = 1/2mV^2, solve for velocity\n\nsqrt[2(9.8m/sec^2*20meters)] = v = 19.8m/sec\n\nb) the work done by the gravitational force on the particle during its motion from P to the ground.\n\nMaybe this is a trick question, I will take it at face value, when at P it is 60meters above the ground, the work of the gravity bringing it down from P is just mgh = 0.5kg*9.8m/s^2*20m=98J\n\nc) the horizontal and vertical components of the velocity vector when the particle reaches the ground:\n\nHorizontal velocity is constant at 30m/sec, nothing to calculate.\n\nVerticle velocity, keep in mind that the object reached its maximum height at 80 meters high (ie from a verticle perspective is is at zero velocity at that point, and begins to fall, the terminal vertical velocity comes from the potential energy of \nmgh = 0.5kg*9.8m/s^2*80m = 1/2 mV^2, solve for v\n\nvertical velocity is 39.6 m/sec downward\n\nplease check for calculation errors.
759319
acceleration=rate of change of velocity\n=80*5280/3600*6ft/sec^2\n=19.6 ft/sec^2\n\n2.0=(70*5280/3600)^2-2f(157)\n314s=67.1ft/s^2
404970
42.43 ft.  60 squared plus 60 squared equals 7200.  take the square root of that and divide it by 2 to get the distance from the center of the diamond to home plate.  its simply the pythagorean theorem.  haha.
485964
what they are doing calculating the speed of a mlb pitch based on the little league dimensions. little league pitchers only throw from 45 feet as opposed to 60.5 feet of the mlb. 74mph from 45 feet is equivalent to 99mph from 60.5 feet. it in no way implies that he throws 99mph. what they want you to see is that if this were a mlb game this is how fast the pitcher would be throwing in comparison. the reaction time is the same for both batters .42 sec
905071
The trick to solving this problem is to set up your equations properly:\n\nR = Vx * t\n\nH = Vy * t - (1/2)g * t²\n\nWhere R is the projectile's range and H is its initial height, Vy is the projectile's initial vertical velocity and Vx is its horizontal velocity.\n\n[[sorry for the edit, I just noticed the 45° angle; that changes the analysis]]\n\nHere's what we'll do: let's flip the problem around and pretend that instead of landing at a downward angle of 45°, it takes off at an upward angle of 45° with some velocity V, and at some point, t, the projectile has the coordinates (50, 40).\n\nThis introduces two new equations:\n\nVx = V * cos(theta)\nVy = V * sin(theta)\n\nCombining the equations now provides:\n\nR = V * cos(theta) * t\nH = V * sin(theta) * t - (1/2)g * t²\n\nSolving for t in equation 1:\n\nt = R / (V * cos(theta))\n\nSubstituting this into equation 2:\n\nH = V * sin(theta) * R / (V * cos(theta)) - (1/2)g * (R / (V * cos(theta)))²\n\nH = R * tan(theta) - (1/2)g * (R / V)² * (1/ cos(theta))²\n\nWe can now solve for V as:\n\n-H * cos(theta)² + R * tan(theta) =  (1/2)g * (R / V)²\n\n[-H * cos(theta)² + R * tan(theta) * cos(theta)²] =  (1/2)g * (R / V)²\n\nV² = (1/2)g * R² / [-H * cos(theta)² + R * tan(theta) * cos(theta)²]\n\nV = [(1/2)g * R² / [-H * cos(theta)² + R * tan(theta) * cos(theta)²]]^(1/2)\n\nSubstituting the values provides:\n\nV = 49.497 m/s\n\nFinally, plugging this back into the expression for Vx:\n\nVx = V * cos (theta)\n\nVx = 49.497 * cos (45°)\n\nVx = 35 m/s\n\nIf we now, flip our coordinate system back around, we know that the horizontal component of velocity is 35 m/s (which does not change over time). However, this implies that there will be some vertical component of velocity when the projectile is fired. We find this by knowing that the final vertical component Vy equals the final horizontal component (to form the 45° angle):\n\nVx = Vy\n\nAdditionally, the vertical component of velocity changes due to gravity. Thus, the initial vertical velocity component is:\n\nVyo = Vy - g*t = Vx - g*t\n\nWhere t = R/Vx = 50/35 = 1.42 s. Plugging this in, we get:\n\nVyo = 21 m/s (downward)\n\nFINAL RESULTS:\n\nVx = 35m/s\n(Horizontal Velocity)\n\nVy(t) = -21 - 9.8*t\n(Vertical Velocity as a function of time)\n\nH(t) = 40 - 21*t-(1/2) * g *t²  \n(Projectile Height as a function of Time)\n\nR(t) = 35 * t\n(Range as a function of time)\nHope that clears things up a bit! :)\n\nVo = sqrt(Vx² + Vyo²) = 40.82 m/s\n(Launch Velocity)\n\nLaunch Angle = arctan(Vyo/Vx) = -30.96°\n\nTime of Flight = 50/35 = 1.42 seconds\n\n-----------\n\nYes, it's long, I know. But I hope it clears stuff up for you :)
333602
If you mean by 'above' you measure along the ramp, then the vertical drop is 40sin(30) which is 20 meters.\n\nSo, you just need to figure out how fast an object will be traveling after falling 20 meters (I am leaving off friction here).
1058066
60 feet, 6 inches\n\nand 90 feet between bases
648817
a) Supposing that s0 = 0. We  have v0 = 32ft/sec\n=> s = -16t^2 + 32 t\n\nb)When t = 1, s=-16 + 32 = 16 ft above the ground\n\nc)When s = 0 again:\n0 = -16t^2 + 32 t\n=>t(-16t + 32) = 0\nThe equation is equal to zero when t=0 (the time the ball was thrown) and when t = 2 => The ball will hit the ground at t=2\n\nd)ds/dt = -32t + 32\nThis is equal to 0 when t=1\nat this point s = 16.
919976
Apparently we are ignoring any pitching motion due to their getting in the car and due to the bump, since you'd need to know a lot more about mass distribution if it was pitching. So we're only considering vertical movement and the distance between axles is irrelevant. I'm assuming "a total of 8.9 cm" means the sum of the 4 spring deflections, so their average deflection is 2.225 cm. If it actually means the average deflection, then k and m are 1/4 what you'll get with the following calculations. Stiffness k (of the 4 springs in parallel) in n/M = 146*g/.02225. Natural radian frequency w = sqrt(k/m), so m = k/w^2 = k/(2*pi/1.7)^2. (That's the total m; mass of car = m-146.) \nThis was not a well-phrased question, because "a total of 8.9 cm" also could mean 8.9 cm from their no-load length (and thus due to the weight of the car and the students). However this would result in an insoluble problem. \nBTW, why are they geeky? Did they figure all this out before me?
74061
the simple answer is if you were to throw a ball at the groud 15 feet away from you the angle created between the ground and the path of the ball when its traveling towards the ground is teh angle of incidence\n\nhope this helps