ballistics, simple formula required to determin how fast an object must be travelling to reach a hight of 200m
362696
The only link you need:\n\nhttp://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html\n\nHint:  look under "Mechanics"
104839
When an object is launched under the influence of only a uniform gravitational force field, it is said to follow a projectile motion. Under ideal condition, the path it traces is parabolic. There is catch here – if the initial velocity is equal or more than the escape velocity then the object will simply vanish into space i.e. leave earth!\n\nImagine if the launch angle from the horizontal is “theta” and initial velocity is “U” then you can easily compute vertical (Uv) and horizontal (Uh) components of it. The Uv will be equal to U x Sin(theta) and Uh will be equal to U x Cos(theta). Now by using Uv you can compute the max height it will reach and the total time it will take to go up and return to ground. With this round trip time in hand, you can easily compute the max horizontal distance it will cover before hitting the ground.\n\nAll these calculation assume ideal conditions like no wind resistance etc.
479843
I'm assuming it's okay to neglect air resistance in this problem, and that the vertocal velocity component you requested is right before the thingy (I love the word "thingy") hits the ground.\n\nWe have the kinematic equation: S=160+V(initial)t+.5at^2.\na is -32 ft/sec^2, and the initial vertical velocity is 0, so we have S=160-16t^2.  When it hits, S=0, so:\n\n160-16t^2=0 --> 16t^2=160  -->  t^2=10  -->t=10^(1/2)\n\n(Or t = (square root of ten) seconds.  About 3.16 seconds.)\n\nNow that we have t, we just multiply that by the horizontal velocity:\n\n960*10^(1/2), or 960 square root of 10.  (In feet.)  \n\nTherefore the object will land 960*10^(1/2) feet from the base of the building.  Roughly this is about 3035.78 feet.\n\nTo get the final vertical velocity, we use the equation:\n\nV(final)^2=V(initial)^2+2a delta S.  \n\nChange in S in the vertical direction is -160 ft, and a=-32 ft/second squared, and V(initial)=0, so:\n\nV^2=2(-32)(-160)=10240 ft squared per second squared.\n\n--> V=32*10^(1/2), or about 101.19 feet per second.\n\nI hope that's correct.  I typed it with one hand while not really paying attention.  ^_^  From now on, do your own homework.  :P
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 :)
378965
975 m/s (3,200 ft/s), 884 m/s (2,900 ft/s) effective to 600 yards. 5.56mm
461302
So far, you have a starting position and an ending position.  If this were a physics question, you'd also need numbers like the mass of the ball and the initial launch velocity.\n\nIn general the equation of motion is given by\n\ns = s0 + v0t + at^2\n\nwhere you need or have the final position (which we have), initial position (which we have), initial speed (which we're looking for, gravity (a known constant), and the time in flight (which we can find using other equations).  Generally, things will go up and down parabolically, not in a straight line, so we need more information about this to answer your question.\n\nNow if you're just looking at a straight line path, then you can use some trig to find at least the angle to the ball (you can't find the speed if you're assuming straight line travel).  In this case, if the person is 1.7 m, and he is standing 5.8 m from the basket that is 3.05 m away, you have a right triangle where the horizontal leg is 5.8 m and the vertical leg is 4.1 m (the difference between the height of the basket and the player, assuming that the ball is thrown from 1.7 m above the ground).  Using the tangent function, you can find that angle:\n\ntan x = opposite side / adjacent side\n= 4.1 / 5.8\n= 0.7069\nx = 35.26 degrees\n\nBut again, laws of motion dictate that the ball follow a parabola to the basket.  But at least we can say that the launch angle has to be at least 35.26 degrees.
786994
velocity-height and horizontal movement. 11.2 km/s in order to escape earth's atmosphere (air provides resistance against movement) above 50 km and movement in horizontal direction so that the object falls perpetually towards the earths gravity centre.
893929
Fg = mg = (0.20 kg)(9.80662 m/s^2) = 1.961324 N\nFc = (mv^2)/r = mrω^2\nFc = (0.20 kg)(0.60 m)((2π rad/s)^2\nFc = 4.73741 N\nFt = √( 4.73741^2 + 1.961324) @ arctan(1.961324/4.73741)\nFt ≈ 5.127 N @ 22.5°\n\n[The vertical component of Ft must offset the weight of the ball.]
214186
Maximum speed: Mach 2.05 at altitude, 1,321 mph (2,100 km/h)
916863
I get that the maximum distance that the ship can fire is 6371m, meaning that any distance greater than 3571m from the western shore should be safe.  It may be tempting to assume that the target ship would be safe within a certain minimum distance, where it would be protected by the mountain peak, but the enemy ship could hit any target within the 3571m limit by angling its guns higher than 45 degrees.\n\nEDIT:\nTo start solving our problem, let's figure out the maximum distance that the enemy ship can fire.  It's not hard, and we know we'll have to figure it out anyway.\n\nIf the ship fires its projectile at exactly 45 degrees, the projectile will make a parabolic arc through the air.  At angles of more than 45 degrees, the projectile will fly higher, but it won't go as far (like a pop-up fly in baseball).  At angles less than 45 degrees, the projectile will fly lower, and hit the water with more horizontal motion (like a line-drive in baseball).  At exactly 45 degrees, it will achieve its maximum horizontal distance.  \n\nIn order to figure out how far the projectile flies, we need to break up its motion into horizontal and vertical components.  Imagine a right triangle with a hypotenuse of 250 units and one angle of 45 degrees.  (The other angle will also be 45 degrees, but that doesn't matter for now.)  The line adjacent to our angle will be as long as the projectile's horizontal momentum.  The line opposite the angle will be as long as the projectile's vertical momentum.  We just need a little bit of trigonometry to figure out these numbers.\n\nsin(a) = opposite / hypotenuse;\nsin(45) = Vv / 250;   (Vv represents the vertical velocity)\nVv = sin(45) * 250 = 177m/s;\n\nSince we're using a right triangle with two 45 degree angles, both sides will be the same length.  If we were using a different angle, we could easily solve for the horizontal momentum using the Pythagorean Theorum.\n\nSo, we know that our projectile had a horizontal velocity of 177m/s and a vertical velocity of 177m/s.  Let's figure out how long it would stay in the air.  We know that it's constantly accelerating towards the earth due to the force of gravity.  Our answer lies in the definition of acceleration: acceleration is a change in velocity over time.  Mathematically, that's (a=ΔV / t).  The "Δ" symbol is the Greek letter delta, often used to represent a change in a value.  \n\nTo figure out how long it stayed up, we only need to worry about the projectile's vertical velocity.  We know that it's initial vertical velocity is 177m/s.  We also know that it's vertical velocity at the top of the arc is 0m/s.  The change between these two points in time is (177m/s - 0m/s =) 177m/s.  The only other thing we need to figure out time is the acceleration due to gravity.  On Earth, it's always 9.81m/s^2. So...\n\na = ΔV / t;\n9.81 = 177 / t;\nt = 177 / 9.81 = 18.04 seconds.  That's how long it takes the projectile to reach the top of it's arc.  It will take equally as long to reach the bottom of it's arc, so the total flight time is 36.09 seconds.  \n\nNow we have a simple distance equation.  Since distance is equal to velocity multiplied by time, the projectile's horizontal distance will be equal to its horizontal velocity multiplied by the amount of time that it stayed in the air.\n\nd = 177 * 36 = 6372m.  (This number is a little higher than the first one, because I rounded a little differently.  It doesn't make a huge difference.)  If we take out the distance to the western shore (2500m + 300M), we get a maximum distance of 3572m from the western shore.  Any ship beyond that distance will be safe.  \n\nAll we need to do now is figure out whether or not the enemy ship can fire at that maximum distance.  If the mountain is too high, the ship will have to increase the angle of its guns, thereby losing range.  We can figure out how high our projectile is going to fly with
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).
815859
the vertical component of the velocity applied to the ball is given by\n\nsin 35 = velocity/24.5\n\nvertical velocity is 14.05 m/s\n\nV^2 = u^2 +2gs\n\nat the maximum height, v = 0 so\n\n0 = (14.05)^2 + 2(-9.8)s\n0 = 197.48 + -19.6s\n19.6s = 197.48\ns = 10.07 metres\n\n\nmaximum height the ball reaches is 10.07 metres...
1163903
Set m1 = 0.145 kg, v1i = 0.720 m/s\nand m2 = 0.295 kg, v2i = 2.22 m/s\n\nConservation of linear momentum:\nm1v1i + m2v2i = m1v1f  + m2v2f\nSimplified\nm1(v1i-v1f) = -m2(v2i-v2f)\n\nConservation of energy (kinetic only):\n(1/2)m1v1i^2 + (1/2)m2v2i^2 = (1/2)m1v1f^2 + (1/2)m2v2f^2\nSimplified:\nm1(v1i - v1f)(v1i + v1f) = -m2(v2i - v2f)(v2i + v2f) \n\nSolving the two simplified equations:\nv1f =((m1-m2)/(m1+m2))v1i + ((2m2)/(m1+m2))v2i\nand\nv2f =((m2-m1)/(m1+m2))v2i + ((2m1)/(m1+m2))v1i\n\nTherefore,\nv1f = 3.315 m/s\nv2f = 1.23 m/s
641828
500ft\nbut you still have to be careful of whats behind your target.
754389
Learn to answer these questions yourself!\n\nIf the air is assumed to be frictionless, the vertical and horizontal velocities are independent.  The altitude is given by H = H0 + V0t + 1/2 a t^2.  H0 is +30 meters, V0 is +24.5 m/sec, and a = -9.8 m/sec^2.  If the target is at ground level, you are solving a quadratic for H = 0.\n\nSolving the quadratic gives you the time of flight, t.  Multiply the time of flight by the horizontal velocity (49 m/sec * cos(30°)) and  you get the distance of the target.\n\nOf course, you can just look at the problem and the answers and see if there's just one answer that's anywhere close.  In this case, the initial vertical speed is 24.5 m/sec; it takes 2.5 seconds for the arrow to rise to its peak altitude, and another 2.5 seconds to come back down to the altitude of the bow (5 seconds total).  Falling the final 30 meters would take on the order of a second.  The cosine of 30 degrees is about 0.87, so the horizontal velocity is about 42.5 m/sec; multiply by 6 seconds, and you get about 250 meters.  Ergo, (4) is the only answer that's even close.
864791
A.) 3.02891 seconds\nB.) 290.1 m/s\n\nTrust me... =)
1135961
in order to solve this problem, you need to work part (b) first before (a)\n\nb): you know gravity (g=32.2 ft/sec) acts opposite the rocket, so calculate, using:\n\nv_final = v_initial - 2*g*y_1\n\nyou know that the final velocity is zero (max height) and you are given the inital velocity. the negative sign is to account for the opposite direction of gravity. move the equation around to solve for y_1 (height from tower to max. height). then add 100 feet (to account for the tower) and you have your answer.\n\na): since you know your maximum height (from part b), substitute that answer for h. [*NOTE: remember to subtract 100 feet from your answer in part b, because t=0 at the top of the tower] You now have a quadratic equation and solve for t using the quadratic formula (you should know what that is). your answer will be the answer!\n\nThis prob is easy only because it worries upto the max height. if we had to worry about the falling down part, well thats another story.\n\n** i already worked out this prob, so if you want to ask more questions/verify your answer, just post an additional response and ill modify my question to answer it. GL!