Question

27). A projectile is launched at a speed of 12 km/s at an angle of 43° above the horizon. Ignore friction for this problem. Where will it be two days after it is launched? Where will it be one year after it is launched? Explain.

Answer 1

Here we apply concept of kinematics and parabolic motion. We apply equation of motion for constant acceleration along horizontal and vertical separately.

Launching epeed, y = 12-km/ecc = 12;000 ms 0 - 436_abore horizontal - Along a-ame Lu x _ initial velocity - Yox = V. Caso accelercetion, Clye = 0 Along yaxle, acceleration, ay = -9.8m/s2 initial velocity, you = y sino (Downward ake taken negdetive). x= voxt + ļ as t2 orxa Vo Cees o # -0 1 Y= vout thay the oya vosinot Źg? ☺ two days after launch += 2 X24X3600 sec =172800 sec y = {12,000) Sin 43°(172800) - 469.89(172800) We can see y co not possible. for vertical displacementy 7=0 will give time of flight To = vo sin o H - 1982 - = 2 Sino - 2C12000 Mb) Sin 430 9. 8 m/s2

ht=1670.2 sec so projectile will be on ground after 16 70-2 sec . - 2 tape after or 1 year later it will be on ground at distance a x= Ų Caso (1670-2) 2x (12000ms) Coos 4.39 (1670-2see >>X= 14658084 meter orla = 14, 658.884 kon do after two days, projectile will be on around at distance 14,658.084 km from Launch point After year also, same distance. 9143658084 keni