Drewser said:
Homework Statement
A rocket is intially at rest on the ground when its engines are fired the rocket flies off in a straight line at an angle of 53.1 degrees abovie the horizontal with a constant acceleration of magnitute g, the engines stop at time T after launch after which the rocket is a projectile Ignore air resistance and assume g is independant of altitude
C) Find the max altitude reached by the rocket andswer in terms of g and T
D) Find the horizontal distance from the launch point to where the rocket hits the ground answer in terms of g and T
(already got A and B CD are the last two)
Homework Equations
Im thinking that you use
y= 1/2gt^2 + Voyt+ Yo
The Attempt at a Solution
y=1/2gt^2 + Voyt+ Yo
y= 1/2 .8gT^2+0T+0
y=1/2 .8gT^2
now i got the .8 from using Voy=vsin53.1 is that correct?
and then after all that I just added that equation to the distance from when the rockets turned off the engines to when the verticle velocity would = 0... can someone please do this problem and explain in detail the best you can for each step and why you did it, i don't even know where to start with D
The initial velocity of the rocket isv_0 v_{0y} is its y-component. The initial velocity of an object that is accelerated
from rest is always zero and so v_{0y}=0.
The equation for the height of the rocket at a time t should be
y=\frac{1}{2}a_yt^2+ v_{0y}t+y_0
not y=\frac{1}{2}gt^2+ v_0yt+y_0
where a_y=gsin(53.1^{\circ}).
In addition, there is no reason to assume that the rocket reaches its maximum height at time t=T; just because the rocket stops accelerating at t=T, does not mean that it suddenly stops moving. Instead, it will begin decelerating due to gravity starting with an initial velocity v_T=v(T) until it reaches a maximum height, at which v_y=0, at some time later than T and then continue falling until it hits the ground.
To calculate this height, you will
first need to determine v_T=v(T) from v_y(t)=a_yt+v_{0y} and
then find the time at which the rocket reaches its maximum height i.e. when v=0 (of course v=0 both at t=0 and at t=t_{max} but you are obviously looking for the t=t_{max}>T solution) and
then finally you can use the equation y=\frac{1}{2}a_yt^2+ v_{0y}t+y_0 to find y_{max}=y(t_{max}). Be careful to use a_y=gsin(53.1^{\circ}) only for the part where the rocket is accelerating upwards (i.e. when T>t>0) and a_y=-g when the rocket stops firing and starts falling due to gravity. Also I strongly recommend you post both the questions for (a) and (b) as well as your solutions, so that we can check them. After you properly work out (a), (b), and (c) I will help you with part (d).
