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sylar· · 1 143 gezek okaldyteswirler.com

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 79.6 m/s at ground level. The engines then fire, and the rocket accelerates upward at 3.90 m/s2 until it reaches an altitude of 1160 m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of −9.80 m/s2. (You will need to consider the motion while the engine is operating and the free-fall motion separately.)

(a) For what time interval is the rocket in motion above the ground?

(b) What is its maximum altitude?

(c) What is its velocity just before it hits the ground?

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Change 31 oktýabr 2010 ý. 09:45 +1
Efendi :
Menden hemmanize plus..! Beytmanhow. Gaharynyz gelse kellanizi stena uruñ.


minusyn name zyyany bar, kop alsan name bolya? Kop goshmak alsan name bolya?

hojik 31 oktýabr 2010 ý. 10:34 +1

fizikden hep 3 alirdim 😆

388@0 31 oktýabr 2010 ý. 11:45

a) t=43.972 s b) h=1944.906 m c) v=195.244 m/s

388@0 31 oktýabr 2010 ý. 11:55
388@0:
a) t=43.972 s b) h=1944.906 m c) v=195.244 m/s

dogrymy jogaplar ozumche chozdum

Greenlaw 31 oktýabr 2010 ý. 15:43
388@0:
a) t=43.972 s b) h=1944.906 m c) v=195.244 m/s

molodes. Ýöne jogaplar ýalñysh ýaly görünýär

sylar 2 noýabr 2010 ý. 06:34
error:
388@0:
a) t=43.972 s b) h=1944.906 m c) v=195.244 m/s
molodes. Ýöne jogaplar ýalñysh ýaly görünýär

gaty dogry error %10'ny geçenok thanx👍

dredman 2 fewral 2011 ý. 16:04
Milan:
Come on guy this is a basic physics question. This is nothing more than kinetics of particles.

totally agree