The range of a projectile launched at an angle is the same for which two angles?
813272
no its wrong\nthe correct answer is \n15 degrees and 75 degrees\nbecause the range of a projectile is same at \nangle A (say) and 90 degrees- angle A
506920
You would agree that, the projectile motion is nothing but a trajectory. It is explained as, the path traced by an object in space.\n\nTherefore when we study the motion of a rocket in space, yes we can call it a projectile.
96240
Use kinematics! Assuming air resistance to be negligible, as it should be for an arrow, the horizontal velocity of a projectile (i.e. an arrow) is constant. Thus, the velocity at which the arrow leaves the bow is constant after it has left the bow. Time the amount of time it takes for the arrow to reach the target after it left the bow. Then, use the fact that v=x/t to find the horizontal velocity of the arrow. Since if the arrow is fired horizontally from the bow, the only velocity imparted to the arrow from the bow is horizontal, the muzzle velocity should be equal to the horizontal velocity that the arrow has in flight. The vertical velocity of the arrow will be constantly changing through out its flight (it will have a constant downwards accelleration due to gravity of 9.81 m/s^2) but, assuming once more that the arrow is horizontal in the bow, there will be no vertical motion in the bow. I suggest using a video camera to time the amount of time it takes for the bow to reach the target after it has left the bow. That way you can use a frame by frame method to get a really precise time of flight. The distance the arrow travels during that time is really easy - it's just the distance between your bow and the target.
496012
javelin thrown at angles 30 and 60 will land in same spot because as per properties of projection if x is angle of projection then range for angles of projection x and 90-x will be same
774526
On a smooth surface, the angle of reflection will be the same as the angle of incidence.\n\nTherefore, the answer is (B) 40 degrees.
1012557
maximum effective range at a point target is 550 meters.  at an area target is 800 meters.  Maximum range is around 3500 meters (meaning you can hit something that far out, but it's a big crapshoot).
72876
He means does the angle of launch (theta) affect the range (R), when you launch an object at some constant speed (u)?\nThis is a basic question in applied mathematics (mechanics).\n\nYes it does.\nWrite the equations of motion for the object.\n(Neglect wind resistance, which is secondary.)\nAssuming the launch surface is flat and you did not launch at an elevation, thus y(0)=0,then the components of velocity in x-,y-directions are:\n u_x = u*cos(theta)\n u_y = u*sin(theta)\n\nNoting the acceleration: a_x(t) = 0, a_y(t) = -g (always downwards), then the velocity components are:\n v_x(t) = u*cos(theta)\n v_y(t) = u*sin(theta) - g*t\n\nHence distance traveled is:\n s_x(t) = u*t*cos(theta)\n s_y(t) = u*t*sin(theta) - 1/2 g*t^2\n\nThe object lands at some time t=t' when s_y(t') = 0 again (not the t=0 solution, obviously). Thus:\n 2*u*t'*sin(theta) = g*t'^2\n 2*u*sin(theta)/g  = t'\n...\nand the range R is the s_x traveled at this time t=t' : \n R = s_x(t')\n   = u*t'*cos(theta)\n   = 2*u^2*sin(theta)*cos(theta)/g\n\nNow u,g are constant here, hence range R=R(theta) is purely a function of theta:\n R = (u^2/g) * 2*sin(theta)*cos(theta)\n R = (u^2/g) * sin(2*theta)          ; [a trigonometric equality that sin(2*theta) = 2*sin(theta)*cos(theta)]\n\nIn this case you can show (without even needing differentiation) that R achieves a maximum when:\n sin(2*theta) = 1\n i.e. theta = 45 deg\n which is what you might have suspected.\n\n============================================================\n\n[For the general solution of launching at an elevation, y(0) = y_0 , then you need to rewrite s_y(t), solve as above, using differentiation to show R(theta) achieves a max when R'(theta)=0 for some theta which does not have a neat solution].\n\n[If you want to be a smartass, add the second-order air resistance terms into s_x(t), s_y(t), so you get acceleration due to a viscous force:\n a_x(t) =      -nu * v_x(t)^2\n a_y(t) = -g + -nu * sgn(v_y)*v_y(t)^2\n   [Note we have to use sgn(y_t) to ensure the viscous forces acts against direction of motion. Basically: down when the projectile is rising, and up when it's falling.\nBut you can usually neglect the air resistance component of a_y, since |v_x|>>|v_y| for most if not all of the flight, unless u is small and launch angle theta is large.]\n\nIf you neglect the air resistance component of a_y(t), then you can still get an analytic solution, otherwise I think now you get two difference equations and you have to solve approximately e.g. by Runge-Kutta approximation.\n\nIf you want to be an utter smartass, add in the effect of wind with a vector (w_x,w_y), such that the air resistance force depends on the relative velocity of projectile to wind:\n( v_x(t)-w_x , v_y(t)-w_y ) \n\nAlso, we neglected terrain (mountains etc.), these degenerate into piecewise tests to make sure you always maintain positive altitude s_y(t).\n\nIf you want to get silly, do the ballistic projectile case, where g decreases as a function of altitude, as does nu, and over long distances you have to factor in the earth's curvature, and probably you want to write it in polar, rather than Cartesian, coordinates.\n g(s_y) = g_0  - k_g(s_y(t))   (linearized approximation)\n nu     = nu_0 * exp (-k_nu* (s_y(t)+Earth's radius)) (atmospheric density)\n [but you can linearize this too, for the lower atmosphere anyway: http://www.aerospaceweb.org/design/scripts/atmosphere/ ]\n\nAnd remember this was just for a projectile, where mass=constant. For a rocket, mass is decreasing (non-linearly) over time, and the dm/dt change in mass is proportional to the acceleration force. The direction of the acceleration is of course along the current velocity vector.\nYou may have design limitations imposed on the maximum dm/dt (e.g. the fuel pumps have a maximum throughput). And you could have two- or three-stage rockets.\n\nIsn't it fun how complicated you can make simple things?\n\nHere's a
703199
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.
447650
Equilibrium is defined as the state of undergoing equal forces in all directions (that is mathematically equivalent to zero force).  A moving projectile, with no gravity acting upon it, and no air resistance (i.e. travelling in a vaccuum) would fit that definition.
572729
When two lines are not perpendicular, the types of angles that can be formed are:\n\n(1) Acute - one of the angles formed by the two lines is less than 90 degrees.\n(2) Obtuse - one of the angles formed by the two lines is more than 90 degrees.\n(3) Straight - the two lines are somewhat inclined in the same direction\n(4) Reflex - one of the angles formed by the two lines is either greater than 180 degrees, whilst the other is less than 180 degrees (which the exception that the two lines form a 270 degree, and a 90 degree angle). This applied to angles of a revolution.
942141
Projectile motion always deals with motion in two different dimensions; the horizontal and the vertical.  Horizontal projectiles always have zero initial velocity in the vertical direction, and nonzero initial velocity in the horizontal direction.\n\nThink of a ball rolling across a table.  While it is on the table, it has no vertical velocity, because the table prevents it from falling.  At the instant that the ball leaves the table, it still has the same horizontal velocity, but has not started moving downward yet.  Since it hasn't started moving yet, its vertical velocity is zero.  Once the surface is no longer beneath the ball, it starts accelerating in the vertical direction, so its vertical velocity would increase until it hits the ground.
479518
The probability would be zero.  Since the angles of a triangle in this case would be considered a continuous random variable (meaning that the angle could be anything in the range of (0, 180)), the probability that you would find a triangle with an angle that is exactly 45 degrees is impossible.\n\nThis is analogous to finding a person who is exactly 6 feet tall.  You may find a person who is close to being 6 feet tall, but their height will always be off by a fraction of an inch.  Same is true with the triangles; you may be able to find a triangle that has an angle very close to 45 degrees, but that angle will always be off by a fraction of a degree.
841719
sorry for laughing but I know everything can be used with math. Those rockets, how long is goes up, where it lands the wind , angle bla/bla
456084
angle b and angle c.\nthey are both on the same line of the x- axis and collide at a central point with respect to A.
42921
optimum angle for distance is 45 degrees \nspeed: fucking fast
1084130
The projectile motion is the motion of a projector starting from origin (x=0 & y=0) with initial velocity (v) & inclination (a) where (a) is the angle between X-axis & (v) so the only accelerations affecting the projector are:\nx''=0 & y''=-g --------> (1) & (2)\n\nTo calculate velocity (x' & y'):\nfrom (1) & (2) we get:       \nx''=dx'/dt=0  \ntherfore: x'=constant=v cos(a) [from initial condition] --------> (3)\n \nfrom (3) we get: y''=dy'/dt=-g\ntherfore: y'=-gt+constant\n\nfrom initial condition:\nat t=0 ----> y'=v sin(a) & constant=0\ntherfore: y'=v sin(a)-gt ------> (4)\n\nThe parametric equation:\nfrom (3): dx/dt=v cos(a)\ntherfore: x=tv cos(a)+constant\n\nfrom initial condition:\nconstant=0\ntherfore: x=tv cos(a) -----> (5)\n\nfrom (4): dy/dt=v sin(a)-gt\ntherfore: y=tv sin(a)-0.5g(t^2)+constant\n\nfrom initial condition:\nt=0 therfore y=0 & constant=0\ntherfore y=tv sin(a)-0.5g(t^2) -----> (6)\n\nfrom (5): t=x/[v cos(a)]          sub in (6)\ntherfore\ny=[x tan(a)]-{ [g(x^2)]/[2(v^2)] }*[1+ tan^2 (a)] \nthis equation is called The Cartizian Equation\n\nTo find The Range ------> sub y=0 in The Cartizian Equation\ntherfor:\nR(range)=[2v sin(a) cos(a)]/g
859137
first thing u should do is do draw the path of the projectile,  using only imagination, the ball will definetly fall down as it moves away from the table edge until it finally hits the floor.\n\nprojectile path is always parabolic, and the direction of the velocity vector is tangent to the path of the projectile at any point.\n\nat the initial state, the velocity vector was pointing left or right according to your assumption, but as the parabolic path falls down, the velocity vector points down with it, when the ball hits the floor the velocity vector will be perpendicular to the horizontal in the downward direction.\n\nso at 45 degrees downward, the velocity vector has two components, on in the horizontal V*cos(theta) and another downward V*sin(theta), to get how high the ball at this instant apply the same laws but use V*sin(theta) instead of V ,  and to find how far from the table use V*cos(theta) instead of V.\n\nhope this helped.