I have a huge physics test tomorrow so can somone plz explain A B, and C?

9. A rocket is launched at an angle of 53.0 degrees above the horizontal with an inital speed of 100 m/s. It moves 3.00 sec along its initial line of motion with an acceleration of 30.0 m/s^2. A this time its engines fail and the rocket proceeds to move as a free body. Find (a) the maximum altitude reached by the rocked, (b) its total time flight, and (c) its horizontal range.


I think I get C but I have no clue about how to solve A and B.

I have a huge physics test tomorrow so can somone plz explain A B, and C?
Sorry I didn't get to this before the test.





You have two situtations, each of which can be described by constant acceleration equations.





Rockets on:


Acceleration a constant 30 m/s^2, in the direction of the initial rocket blast, which is the same as the line of motion. Acceleration is a vector, and has x and y components:





ax = 30 cos 53 = 18.05 m/s^2


ay = 30 sin 53 - 9.8 (the object is pulled by gravity, even when the rockets are firing) = 14.16 m/s^2





The initial velocity components are


vx = 100 cos 53 = 60.18 m/s


vy = 100 sin 53 = 78.86 m/s





v(t) = a*t + v0. Let's evaluate this to get the velocity in each direction after 3 seconds





vx(3) = 18.05(3) + 60.18 = 114 m/s


vy(3) = 14.16(3) + 78.86 = 122 m/s





x(3) = (1/2)18.05 * 3^2 + 60.18*3 = 262 m/s


y(3) = (1/2)14.16 * 3^2 + 78.86*3 = 303 m/s





These final velocities will be the initial velocities for the motion after the rockets fail.


The new equations are:


x(t) = (1/2)*18.05*t^2 + 114*t + 262


y(t) = (1/2)(-9.8)t^2 + 122*t + 303





vx(t) = 18.05t + 114


vy(t) = -9.8t + 122





A. Maximum altitude:


This occurs when vy = 0.


Solving 0 = -9.8t + 122 for t gives t = 12.5 s.


Now plug this time into y(t)


y(12.5) = -4.9 ( 12.5)^2 + 122 (12.5) + 262 = 1066m.





B. Total time in flight.


The flight ends when the rocket hits the ground.


Mathematically, -4.9t^2 + 122t + 303 = 0.


The positive solution to this equation is t = 27.24. The total time of flight is 30.24, including the initial 3 second engine burn.





C. Horizontal range


Now we can place the time of flight (not counting the initial 3 seconds) into


x(t) = (1/2)*18.05*t^2 + 114*t + 262 = 10074.83





It's been a while since I've done this. Hope this helps!
Reply:Since it moves along it's original line of motion, g must be cancelled by part of the upward component of the roket's thrust, and 30.0 m/sec^2 may be considered the only acceleration operating:


v = v0 + at, but because you need to have the vertical and horizontal distances travelled in the 1st 3.00 sec, as well as the velocity, separate the problem into 2 problems:


Vx = V0x +AxT, X = V0xT + (1/2)AxT^2, and


Vy = V0y +AyT, Y = V0yT + (1/2)AyT^2





Vx = 100cos53 +30(cos53)(3) = 190cos53


X = 100cos53(3) +(1/2)30(cos53)(3)^2 = 435cos53


Vy = 100sin53 +30(sin53)(3) = 190sin53


Y = 100sin53(3) +(1/2)30(sin53)(3)^2 = 435sin53





(Actually I could have waited 'til here to resolve the components--I'm a bit rusty) Now you have the initial conditions for the free-fall portion of the flight.


a) maximum altitude:


2a(s-s0) = v^2 - v0^2, s = (v^2-v0^2)/(2a) + s0





Y = 435sin53 -(190sin53)^2)/(2*(-9.8))


Y = 347.41 + 1174.76


Y = 1522.17 meters





b) total time of flight must also be broken into 2 parts: time to reach zenith, and time to fall back to earth.


Rising time:


T = (Vy - V0y)/A + T0


T = (0-190sin53)/-9.8 = 15.484 sec + 3 sec


Falling time:


2Y = 2V0yT + AyT^2, but v0 is 0 this time.


T = sqrt(2Y/A) = sqrt(2*1522.17/9.8) = 17.625 sec.


T(ff) = 17.625 + 15.484 = 33.109 sec.


T(total) = 36.109 sec.





c) horizontal range


X = V0xT + (1/2)AxT^2 + X0


X = 190cos53)(33.109) + 435cos53


X = 6725.71cos53


X = 4,047.6 meters
Reply:Erm - how can you get its horizontal range without knowing time of flight?





How know its flight time without calculating maximum height?





Very best of luck - Mike
Reply:I cud do this question but iv got november exams cumin up an i havnt done a tap. so here it is for a) the max altitude is calculated by putting the velocity on the y(vertical or J) axis equal to 0. Every thing else shud be pretty straight forward