A model rocket is launched straight upward with an initial speed of 30.0 m/s.?
(a) Use the kinematic equation Vyf^2 = Vy0^2 + 2*a*H to determine Vyf, the velocity at the point where the engines stop: Vyf = sqrt(Vy0^2 + 2*a*H) = sqrt(30.0^2 + 2*1.50*100) = 34.64 m/s Now apply the same equation, only now we are solving for H, the additional height obtained with Vyf = 0, and Vy0 = 34.64 m/s: Vyf^2 = Vy0^2 - 2*g*H = 0 H = Vy0^2 / (2*g) = 34.64^2 / (2*9.8) = 61.22 m Add this back to the original height: Final height = 100 m + 61.22 m = 161.22 m (b) First calculate the time for the rocket to reach the height at which the engines stop. Vf = V0 + a*t t = (Vf - V0) / a = (34.64 - 30.00) / 1.50 = 3.09 s Now calculate the remaining time with the same equation: Vf = V0 - g*t t = (0 - 34.64) / (-9.8) = 3.53 s Add the times together to get the total time: 3.09 s + 3.53 s = 6.62 seconds (c) Now calculate the time required for the rocket to fall from rest back to the earth. Remember the peak height is 161.22 m (from part a), and V0 = 0: H = V0*t + 0.5*g*t^2 = 0.5*g*t^2 t = sqrt(2*H/g) = sqrt(2*161.22/9.8) = 5.74 s Add this to your answer from part b to get the total time in the air: 6.62 s + 5.74 s = 12.36 seconds Hope that helps!
A model rocket is launched from rest with an upward acceleration of 6.00m/s^2?
a model rocket is launched from rest with an upward acceleration of 6.00m/s^2 and due to strong wind, a horizontal acceleration of 1.50 m/s^2. How far is the rocket from the launch pad 6.00 s later when the rocket runs out of fuel? I got 27 m for the final x position but the book's answer is 111 m. The book's answers tend to be wrong from time to time and I cant seem to figure what I might have done wrong so this is my last resort. Can someone verify my answer or not.
A ball is thrown upward with an initial velocity of 30 m/s. what is the maximum height that the ball will reach?
All answers I've read just give you an equation and fill it in. I just don't think that helps anyone understand what is being asked. So here goes my attempt at teaching anyone interested.There actually are several ways to calculate this of which I pick the one that is simplest to me.Assuming there is no (significant) air drag no energy is gained or lost; the kinetic energy (Ek) the ball has leaving your hand is converted in potential gravitational energy (Ep) on the way up, and back to kinetic energy on the way down.Their size must therefore be equal:Ek = EpAlso we know that,Ek = 0.5*m*v^2. and, Ep = m*g*hwith m the mass of the ball, v its velocity leaving your hand, g the gravitational acceleration and h the maximum height the ball reaches.Rearanging a bit gives you:v^2 = 2*g*h or,v^2/2g =hCurious about a different way?Take,St = S0 + V0 *t + 0.5*a*t^2which gives the position of an object at time t (St) given the position at time 0 (S0) the initial velocity (V0) and the acceleration (a) which is constant.For a we fill in -g (because down is negative) we take S0 = 0 and we know V0.To find t we need to solve it from,Vt = V0 + a*t.Since the velocity at the summit of the balls flight is zero we get:V0 = g*t. or, t = V0/g(Notice how I cleverly got rid of a minus?)Fill in that value into the equation for St and you're done.But you can also plug the formula for t into the equation:St = V0*V0/g - 0.5*g*(V0/g)^2= (V0^2)/g - 0.5*g*(V0^2)/g^2= (V0^2)/g - 0.5*(V0^2)/g= 0.5*(V0^2)/g = (V0^2)/2gwhich is the result we had before when you consider that v there is V0 here.Isn't it great how everything lines up?I hope this helped you understand rather than get the answer right.Yes… There's a difference…
A toy rocket is launched vertically from the ground on a day with no wind.?
This equation appears to be wrong . Correct it and someone maybe answer your question correctly v (t ) = 500 ° 32t feet/sec. This implies your rocket only get faster and faster as time goes by. If so, it never will have velocity 0. It eventually reaches near the speed of light, then call Einstien. Is the "°" meant to be a "-" -32 ft/second is the deceleration due to gravity. Therefore , normally v(t) v(t) = v_i - 32t ft/sec where v_i is the initial velocity. ==== Rest of the work is based on assumption of what you're equation should be if the equation was v(t) = 500- 32t then, v = 0 when 500-32t = 0 32t = 500 t = 500/32 = 15.625 seconds What happens to the rocket at t = a? The rocket reaches velocity 0. Then, The rocket begins to accelerate toward the ground based on gravity. b) Find the value of the total area enclosed by y = v(t) and the t-axis on the interval 0 less than or equal to t less than or equal to a. What does this area represent in terms of the physical setting of the problem? Integral of v(t)dt = distance traveled by the rocket . so if you meant v(t) = 500- 32t Integral of v(t) from t to 0 = 500t -(32/2)t^2 +C - (500*0 - (32/2)0^2) - C distance travelled = 500t -(32/2) t^2 = 500t -16t^2 s(t) = 500t -16t^2 + C but if you say s(0) = 0 (initial height = 0) then s(t)= 500t -16t^2 d) Compute the value of s(a) ° s(0). What does this number represent in terms of the physical setting of the problem? what does s(a) ° s(0). (multiply or what ) I don't what multiplying s(a) times s(0) means e) Compute s(5) -s(1). What does this number tell you about the rocket’s flight? This is the distrance traveled by the rocket between 1 and 5 seconds of it's flight. Since the rocket is only traveling straight up to begin with, the distance it travels is how much it rises in altitude. s(5) -s(1) = 500t- 32t^2 = 500*5 -16 *25 - (500 -16*1) = 4*500 - 16(1-25) = 2000 - 16*24 = 1616 feet so the rocket raised 1616 feet in this 4 seconds period.
How do I find max height when only given an initial velocity?
By noting the vertical velocity and noting the equivalence of kinetic and potential energy.Potential energy ismass * acceleration * height = Potential Energy = mghKinetic energy is(1/2) * mass * Velocity squared = Kinetic energy = (1/2) * m * V^2So, setting one energy equal to the othermgh = (1/2) m * V^2We can cancel mass, so it doesn’t really contribute anything thereforeg * H = (1/2) * V^2Now this V is the vertical velocity. If you’re pointing straight up - 90 degrees from horizontal, ALL of your velocity is straight up. Since sine of 90 degrees = 1 we writeg * H = (1/2) (sin(theta) * V)^2theta is called the elevation angle.So, if we shoot a ball across the floor with no vertical velocity, it has zero height. If we shoot a ball straight up, 90 degrees to the floor, then all of the velocity goes into potential energy. If we are at 45 degrees - 70.71% of the velocity goes up and 70.71% of the velocity goes sideways - and we square that and see that 50% of the energy goes up and 50% of the energy goes sideways.Okay, so let’s say we shoot a ball off at 45 degrees at 100 kph. That’s 27.78 m/sec. That’s 19.64 m/sec up and 19.64 m/sec sideways. We are only interested in the up part. I hope you can see that if we fired off at 60 degrees less speed would go into horizontal movement and more into vertical movement. Likewise if we fired off at 30 degrees more speed wold go into horizontal movement and less into vertical movement.Okay at 100 kph and 45 degrees the object is moving upward at 19.64 m/sec so,g * H = (1/2) * (19.64)^2 → H = (1/2) * (19.64)^2 / 9.81 = 19.66 m.Nowvelocity = acceleration * timeand acceleration is 9.81 m/s/s and velocity was 19.64 m/sec so we can see that the time it takes to rise to this height ist = 19.64/9.81 = 2.00 seconds.Now, the horizontal velocity is 19.64 m/sec as well. So, in 2 seconds the object moves horizontally 39.28 meters. Its now 39.28 meters down range and 19.66 meters altitude. Its velocity is zero vertically, but ignoring air drag, horizontal velocity is still 19.64 m/sec. So, it will fall back to Earth - taking another 2 seconds to do that. Moving another 39.28 meters down range. A total of 78.56 meters! (257.68 ft range 64.48 ft altitude)
What happens when rocket is launched with speed less than escape speed of earth?
Escape velocity on the surface of the Earth (about 11 km/sec) is the velocity required for escaping Earth if no additional thrust is generated after the moment it leaves the surface. If you would throw a ball straight up, it would need to go that fast in order to never come back, if you can’t manage to throw it up that fast, it may get very high, but it will always fall back. (I am not taking air-resistance into account, in order to keep it simple). A rocket doesn’t need 1) to escape Earth, just get in orbit, which you can think of to be a balance between gravity and centrifugal force (which doesn’t really exist, but again: we’re keeping it simple), and b) to have escape velocity, since it will keep burning it’s engine, and this way will keep accelerating…You should see Escape Velocity as a characteristic of your situation, not something you actually need to achieve…