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by Jahono on November 24th, 2009
Eqn: d = 1/2 ( Vf + Vi ) × t
As long as acceleration (or decceleration) is constant.
by Cyanotic Wasp on November 24th, 2009
There is insufficient data to answer this question. For one thing, the value for acceleration isn't given, which is an absolute requirement. An object that is "shot" is assumed to have started from a velocity of 0, so what was the initial acceleration used to achieve the 29 m/s value?
Second, there is no launch angle given (I suppose we could assume 90° perpendicular to Earth's surface), and this is also required.
Finally, the "place" of the launch is not mentioned. If this is in Earth's atmosphere, from sea level, then that's one thing. But if it were shot from the Moon, or from the floor of the ocean, then the values would be completely different.
by zeds_dead_baby on November 24th, 2009
well, 29 metres per second.....3 seconds.... pretty obvious isnt it? do your own homework
by Seanzie is smiling again on November 24th, 2009
29 multiplied by three
When did Isaac Newton make the three laws of motion?
by Answerbag Staff on May 19th, 2010
| 1 person likes this
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by Inddiggo:)) on June 23rd, 2011
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| 1 person likes this
Roco = rate of change of. Roco distance = speed. Roco speed = acceleration Roco acceleration = jerk. Roco jerk = jounce. Roco jounce = ?
by Ombliss22 on September 6th, 2011
| 1 person likes this
You're reading An object is shot up at a velocity of 29m/s and it takes 3 seconds to reach its maximum altitude. What is the objects maximum altitude?
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