Question 1087358: An astronaut on Mars throws a ball into the air with an initial velocity of 30 feet per second from a platform 36 feet tall. Gravity on Mars is approximately -12 feet per second per second (-12). So, the height h of the ball t seconds after it is thrown is given by h6t2 30t 36. (a) How high is the ball after 3 seconds?
Found 2 solutions by ikleyn, jorel1380:
Answer by ikleyn(52790)   (Show Source):
You can put this solution on YOUR website! .
1. The Mars atmosphere can hardly be called "the air".
See this description of the Mars atmosphere from https://www.space.com/16903-mars-atmosphere-climate-weather.html :
"The atmosphere of Mars is about 100 times thinner than Earth's, and it is 95 percent carbon dioxide. Here's a breakdown of its composition:
Carbon dioxide: 95.32 percent
Nitrogen: 2.7 percent
Argon: 1.6 percent
Oxygen: 0.13 percent
Carbon monoxide: 0.08 percent
Also, minor amounts of: water, nitrogen oxide, neon, hydrogen-deuterium-oxygen, krypton and xenon."
2. The condition does not say if the ball was thrown vertically up or down, or either other way.
The equation given also does not shed light on this subject.
3. In the future, do not write equations without using signs. Please. Thanks.
4. Still, if your equation is
h(t) = -6*t^2 + 30t + 36, (which means that the ball was thrown vertically up)
then, to answer your question, simply substitute t= 3 into the right side and calculate h(t).
Answer by jorel1380(3719)  (Show Source):
You can put this solution on YOUR website! If the gravity on Mars is equal to -12 ft./sē, then the formula for a projectile, launched from a platform 36 ft. high, with a velocity of 30 ft./sec., would be:
h(t)=-12/2 tē+30t+36
After 3 seconds, you have:
h(3)=-6(9)+90+36=72 ft.
The ball will return to Mars at:
0=-6tē+30t+36
6tē-30t-36=0
tē-5t-6=0
(t-6)(t+1)=0
t=6 seconds before the ball hits the Martian surface.
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