Question

a Rocket launched with initial conditions of: 1) launch velocity = 200 m/s and 2) launch angle= 60 degrees

Answer you should recive for distance and max height

Distance

will land 3600m away

while achieving a

maximum height of 1528m.

Solve for below with LABVIEW

1. Create a LabVIEW VI that will model the launch trajectory that contains the following:

a. Input Numerical Indicators

i. Launch Velocity

ii. Launch Angle

b. Output Numerical Indicators

i. Height

ii. Distance

iii. Max Height

iiii. Impact Distance

c. Output Status Indicator

i. Both Height and distance criteria have been met

d. Output x-y Graph

i. Distance vs Height

Answer 1

The horizontal distance traveled at any time t is:

x = ucos60 t = 100t

and vertical distance traveled in the same time t is:

y = usin60t - (1/2)gt² = 173.2t - 4.9t²

The time of flight of the rocket is:

in this time, the horizontal distance traveled = Range of the rocket

This is, R = 100(35.348) = 3534.8 m

the time when the rocket is at its highest point is, t = T/2 = 17.674 s

so, the maximum height achieved by it is: y = 173.2(17.674) - 4.9(17.674)² = 1530.611 m

For time 0 < t < 35.348s, determine x and y for every time step and plot y vs x graph. The equation of this graph will be:

.