Average velocity of projectile?
Average velocity is distance divided by time. Between two instants, when projectile crosses half the maximum height, it's displacent is zero in vertical direction (from +Hmax/2 to +Hmax/2), and its motion in horizontal direction is uniform. Answer: (Uxavg, Uyavg) = (u cosθ, 0) Uavg = u cosθ
What is the average change in velocity during the total projectile motion of an object?
Simple.If you haven't brushed up on your vectors in a while, now's the chance; Chegg.comLet the initial velocity be u, the final one be v.Let u = xi + yjThen, as Kinetic Energy is conserved, and no acceleration acts along i:v = xi - yj∆v = v - u = 2yj|∆v| = 2y
What is the launch speed of a projectile that rises vertically above the Earth?
This is solvable by finding the difference between the Projectile's gravitational potential energies at Earth's surface and at a height above the surface of six times Earth's radius, then solving the formula for kinetic energy in terms of velocity for that amount of energy. Gravitational potential energy is given generally by U = -GMm/R U = potential energy, in joules G = gravitational constant, 6.67 x 10^-11 M = mass of Earth, 5.97 x 10^24kg m = mass of body at some distance from Earth's center of mass R = distance of body from Earth's center of mass Since all we want is a velocity, and gravity accelerates all masses equally, we can define the value of m to be equal to exactly 1, and the equation becomes U = -GM/R For U at Earth's surface, we'll let R = r. The difference between the two potential energies is GM/r - GM/R r = 6.39 x 10^6m R = 4.47 x 10^7m The difference comes out to 5.34 x 10^7J Kinetic energy is E = mv^2/2. Solving for v this comes to sqrt[2E/m] = v Since the projectile mass is defined to be 1kg, this reduces to sqrt[2E] = v v = 10,334.41 meters per second Note: there is a test for the plausibility of this answer. Since the projectile comes to a halt at some finite distance, its initial velocity must be less than the escape velocity of 11,200 meters per second.
What is the projectile speed and rocket speed in rocket launching?
In rocket launching there is no projectile speed.As projectile means a body acted upon only by gravity whereas rocket is acted upon by gravity as well as thrust generated by fuel combustion.
A particle is projected from the ground with an initial speed of v at an angle Φ with horizontal. What is the average velocity of the particle between its point of projection and highest point of trajectory?
All the answers are incorrect here is the pick of. Solution
How high would the projectile go and what is the magnitude of the average force?
With no air resistance the effect of gravity would be the same regardless of the mass of the object. Find the time it takes to reach max altitude: v = g*t .... t = v/g = 25/9.8 = 2.55 seconds Now the distance: s = (1/2)g*t^2 = (1/2)(9.8)(2.55)^2 s = 31.89 meters So the projectile would reach an altitude of 31.89 meters One way to get the effect of air resistance is to determine what additional acceleration over gravity will now make it only reach this lower altitude and then assign that acceleration to the effect of air resistance. v = (g + a)*t and t = v/(g + a) s = (1/2)(g + a)*t^2 s = (1/2)v^2/(g + a) g + a = (1/2)v^2/s a = (1/2)v^2/s - g = (1/2)(25)^2/7.47 - 9.8 a = 32 m/s^2 And the force is then the mass times this so: F = 0.639*32 = 20.47 N Another way is to balance energy. At time 0 it is all kinetec: E0 = (1/2)mv^2 At max altitude it will all be potential: Ep = m*g*h The change in energy is the difference of these: Echange = (1/2)mv^2 - m*g*h And this can be attributed to the work done by friction: F*h Echange = E0 - Ep F*h = (1/2)mv^2 - m*g*h F = (1/2)mv^2/h - m*g = m[(1/2)v^2/h - g] = 0.639[(1/2)(25^2)/7.47 - 9.8] Force = 20.47 N
Physics problems...help please! (average acceleration and projectile motion)?
Average speed is just the rate of change between total distance and total speed: v=total distance/total time So, we have first to obtain the time the object takes, in each case. Solving for t: v=d/t d=vt t=d/v we obtain, for the case that v=80.0m/s and d=1000 m, t=1000m/(80m/s)=12.5 s and for the case that v=50.0 m/s and d=1000 m, t=1000m/(50m/s)=20s Therefore, total distance is 1000+1000m=2000m and total time is 12.5+20s=32.5s. Therefore, average velocity is: v=2000m/32.5s=61.5 m/s For the second case, we don't care about the y component of the speed, because we're looking for the horizontal distance x, that represents the distance from the table (not the height). In x, we have movement with constant speed if we ignore the air resistance: v=Δx/Δt and we know that v=.24m/s and Δt=0.30 s. We solve for Δx: v=Δx/Δt Δx=vΔt =(.24m/s)(.30s) =0.072m =7.2 cm far away from the table. Note that, in this case, you don't need the value of the gravity because you're only evaluating the x-component of the projectile. When you solve a projectile motion, always consider it as two separate kinds of motion: one in x-direction, with constant speed (accel. 0) and one in vertical direction, with constant accel. g=9.81 m/s²
The velocity of a projectile equals its initial velocity added to: (multiple choice)?
The velocity of a projectile equals its initial velocity added to: A. a constant horizontal velocity B. a constant vertical velocity C. a constantly increasing horizontal velocity D. a constantly increasing downward velocity E. a constant velocity directed at the target
A football is kicked with an initial velocity of 64ft/s at a projectile angle of 45degrees...?
The range (maximum horizontal distance) travelled by a projectile is given by the formula R = V^2(sin 2A)/g where R = range V = initial velocity of ball = 64 ft/sec. (given) A = angle of launch = 45 degrees (given) g = acceleration due to gravity = 32.2 ft/sec^2 (constant) Substituting values, R = 64^2(sin 2*45)/9.8 R = 127.20 feet Since the receiver is 60 yards away (180 feet), he will have to travel a distance of 180 - 127.20 = 52.80 feet to catch the ball. The ball's total travel time is given by the formula T = 2V(sin A)/g where all the terms have been previously defined. Substituting values, T = 2(64)(sin 45)/32.2 T = 2.82 sec. Therefore, in order for the receiver to catch the ball, his speed must be equal to 52.80/2.82 = 18.72 ft/sec. Hope this helps.