Question

Assume a rocket is launched from a platform that is 20 meters high at angle of 53 degrees. While the rocket still has gasoline, it uniformly accelerates, maintains a launch angle relative to
the ground, for four seconds and reaches a velocity of one hundred meters per second starting from rest. You may neglect any air resistance.

Part 1: What is the max height the rocket can have before it falls back to ground?

Part 2: How much time does the rocket spend above ground?

Part 3: What is the distance the rocket covers horizontally?

Part 4: Draw a graph that represents x vs y

Part 5: Draw a graph of x-distance versus time

Part 6: Draw a graph of x-velocity versus time

Part 7: Draw a graph of x-acceleration versus time

Part 8: Draw a graph of vertical position versus time

Part 9 Draw a graph of y-velocity versus time

Part 10: Draw graph of y-acceleration versus time

Answer 1

20mis and © In the problem Vo 53° gi = 9.8mis? a) the height of projectile is given by Component y and it reaches its max. Value when the component Vy is equal to zero. Vy = Vo sim(o) - gt = the formulas for the t = Vosinco) lg components Vx & Vy of velocity and t = 20 Sim (53) / 9.8 V x = Vocoso 2ox 0 79 8 6 Vy = vosimo - gt x = Voloscoot t y = vosince)t- & g +² - 0 dis. x4 y t = g. = 15.972 g. 8 t = 1.63 sec For max. ax. height y (1.63) = 20 Sim (530) (163) -Źx9.8 X{1.(3) y = 20x 0.7906 X1.63 - 4.9 x1-63x1.63 - 26.03 43 - 40g 13.0188 13.0155 13.02 meters y touches the b) the time of flight is the Interval of time b/w when it is launched it, & when the projetile At t=t, & t= tz ry hence Vosincolt- & gt2 = solve for (t) ground tz