my son wants to achieve maximum distance with his remote control car when ramp jumping. im ready to build the ramp but would like to know the optimum angle....
121815
45 degrees is the optimal angle to launch a projectile for maximum horizontal displacement.
625972
This problem is already solved. You know about those guys doing "base jumping"?\n\nTo really answer your question I would suggest a reinforced sphere with a chair in the center contained in an inner reinforcement cage which is attached to the surface of the sphere with springs (in all directions). You have to be very well strapped in the chair (maybe like in a high end  rollercoaster).\nI'll make some calculations but I think the diameter of the sphere should be kinda big (to have a longer stoping distance) in order to avoid high G values
355456
For launch velocity\nv0 = 4.47 m/s,\nlaunch angle\nθ = 66 degrees,\nThe horizontal range is\nR = 1.515173718712374 m.\nThe total time of flight is\nt = 0.83337718278825 s.\nThe peak height is\nh = 0.8507839727702984 m.
424569
Oscar. You have written too much. This 125m tower stuff does not matter compared to r=6.38*10^6. Why is this talk about 1.00 km and angle theta? See what “zipper” above says. Besides I answer your last question << Why is the trajectory of the cannonball not parabolic?>>. Why should it be if both you and Newton demand it to be circular? Assume that V1 of cannonball is the 1st cosmic speed corresponding to a circular track. Let V2 be > than V1, then you receive an oval track. Take V3 > than V2, then the track may become parabolic, so that cannonball would never return back to the cannon to linger out among planets and stars. If you are intrigued calculate a set of velocities and corresponding orbits of the cannonball. If cannon is not available consider a football! I like your inventive wits – to be honest!
333168
Let's consider the case when the Car is right in front.\n\nAt that instance, the car is travelling tangentially to a circle of radius 132ft, where the Camera is the centre of the circle.\n\nCircumference of the Circle = 2*(pi)*R\n= 2 * pi * 132 = 264*(pi) ft.\n\nTime taken to travel the entire circumference by the car\nT = circumference/velocity = 264*(pi)/264 = pi seconds\n\nHence the car would require pi s to cover 360 degrees of the circle.360 degrees = 2(pi) radians.\n\nW = angular velocity = 2(pi)/pi = 2 radians/sec.\n\ncan't help about half a sec later now
528756
Why yes. Secure a tow rope around it and the other end of the rope to the bumper of your best friend's car. Have best friend sloowwwly drive forward (might feel slight discomfort) until you reach desired length. Beware: it is imperative your friend does not crank up the radio during this procedure. He needs to be able to hear your screams.
433053
wish I could solve em for you but ill try and help still.\n\n1) The acceleration should be the derivative of the equation for the car's velocity.  The original distance/time equation would be the anti-derivative of the velocity.  \n\n2 The equation to find out how long it will take to get there is h+ax-16x^2 in feet.  (a is the original speed and x is number of seconds and h can be the height of the building.)\n\nHope this helps
400218
The height of the ramp give potential enery in the following format\n\nE = mgh      where E is potential energy, m is the mass of the car in Kgs, g is the gravitational acceleration of the planet you're on (in meters per second per second (9.8 on earth)) and h is the height of the ramp.\n\nThe angle or slope of the ramp will not affect the potential energy, but will have an effect on the rate of change of potential energy.
99555
Go with 90 degree angle-iron that is set at 45 degrees from vertical.  This will give you a left track that looks like:\n\n>   <\n\nThen you set up your coaster car wheels in the horizontal plane so they are like this:\n\n- >    < -\n\nRiding in the vee of the angle-iron.  This will provide your coaster with both lateral as well as vertical grip, assuming that your angle-iron is sized large enough.
105445
There is a gadget that is made called a launch controller that is made especially for boosted and NOS cars that will get you off the line sooner. Try doing a search to see if you can find one that suites your car. You can also try finding a different gear ratio if you haven't already. Give you quicker traction. Also play around with your tire pressure as well.
233617
I'll get you started...\n\nFirst, determine how long the ball was in the air.  From that, determine the velocity at which the ball was thrown (the rate necessary, factoring in gravity's acceleration, to keep it airborne for that period of time).  Construct a vector using the velocity of the truck vs. the velocity of the ball to determine the angle.\n\nGood luck!
534206
a) 30*5280/3600 *ft/sec = 44 ft/sec\n\nat 150 feet away it will take you 3.4 secs to reach the child\n\nb) 20*5280/3600 *ft/sec = 29.3 ft/sec\n\nat 150 feet away it will take you 5.1 secs to reach the child\n\nc) 10 miles/hr would give you 15 ft/sec and that is a reasonable reaction time to process the problem and react.  That would give you well over 10 seconds to apply the brakes in an emergency.  Plus it would give you better control over the car.
467366
Use the overhead style they teach.  The advantages are that the arc will take it over obstacles like walls, and also when it hits, it won't roll past your target.
658295
It has been proved that the maximum range is obtained when the angle of throw is 45 degree to the horizontal, considering that there is no air resistance.\n\nNow whatever be the angle of throw, the air resistance will be the same. In other words the air resistance is independent of the angle through which it is thrown.\n\nTherefore, with air resistance, the range will be again the maximum if the angle of throw is 45 degree.\n\nIf there is wind depending upon the direction of wind and its speed, he range will be affected. However the range will be the maximum at 45 degree even if there is wind.
79957
Sticking by the rules you specified;\n\nConstruct an inclined channel (ramp) where the 1 kg (2 lbs)weight (hopefully a metal ball) falls and accelerated by gravity. place the car that will catch and hold the ball at the end of the ramp. The falling ball's inertia will propel the car almost equal to its speed and may reach more than 1 meter on a flat even surface, design the car with low center of gravity, with less than half the weight of the propellant ball (weight) as a consideration.\n\n************* \n\nAh, New data acquired --- resolution: same application of inertial force using a rigid pendulum.\n\nConstruct a pendulum with the required height putting the fulcrum point at 10 cm directly above the floor of the car, giving you a swing radius of 10 cm from top to bottom, use a cardboard/plastic tube (shute) to carry the pill bottle but loose enough to set it free after a sudden stop once it reaches the bottom where the position of the pill bottle is parallel with the floor of the car. This configuration may not give you as much distance relying on gravity alone but may be able to get 1 meter if conditions are favorable, pertaining to weight of car and means of motion (wheel and shaft should be well oiled for free-wheeling), relevant to weight of falling object, unless you are allowed to spring load the fulcrum. You can put the fulcrum as part of the car (but it will add more weight/ but you wont need the shute, all components will be integrated), or as a separate launcher (which you will need the shute). make sure the center of gravity on the car is low for stability. when integration of all component is involved make the wheel base equally wide as 10 cm. If just for the pill bottle, just narrow enough to carry the bottle.\n\nmy idea: http://server4.pictiger.com/img/124010/other/1-kg-inertia-car.jpg\n\nInnovation is part application and part logical creativity. Isn't science great!\n~meow
202264
The movie trailer for X3 has a pair of vertical angles in the X.  Also, the Volkswagen kinda has vertical angles (although the vertices aren't shared, so technically it isn't a pair of vertical angles, but it looks close).  Ads for Exxon will work too.
345400
It is an exercise in elementary calculus to show that (ignoring air resistance) the elevation angle for maximum distance is 45 degrees.  The vertical velocity of the egg will be zero at the peak height, and will increase as v = gt^2/2 until it hits the ground.  Energy is conserved, so the vertical velocity when it strikes the ground will match the vertical velocity with which it was launched, and the horizontal velocity will be constant.   This is enough data to solve the whole thing, but there is a shortcut: in the parabolic path at issue, it is always true that the peak height is 1/4 of the horizontal distance.  Hence, the peak height here is 24.225 meters, and that makes it trivial to compute the time and hence the velocity.