Calculating Moon's Gravity: Exploring the Equation h(t) = -.8t^2 + 20t

[lizzygurl88]

Suppose a ball is thrown upward from the surface of the moon with an initial velocity of 20 m/sec. Then its height 'h' meters above the surface is a function of time 't' seconds after being thrown is modeled by the following equation: h(t)= -.8t^2 = 20t

What is the acceleration due to gravity on the surface of the moon? (please show your work or explain how you got your answer) (the correct answer is aboue 1.6 m/s^2...but I need to know how the answer is gotten) THANKSSS!

 

[assyrian_77]

Well, why don't you show some of your work? How are you approaching this problem? And what do you mean by $$h(t)=-0.8t^2=20t$$?

 

[lizzygurl88]

This is the given equation for the problem...it is a quadratic model for the problem, and -.8 is gravity on the moon. (this equation is specific the the question but derives from h= -1/2gt^2 + vo(initial velocity) +ho (initial hight)...i would show work but i am not a physics student..this is for a math class, and I do not know where to start

 

[assyrian_77]

...quadratic model for the problem...

Are you sure that there is an equal sign between the two expressions?

Anyway, you say that the question asks for the acceleration due to gravity on the surface of the moon, right?

Read those words carefully. What do you think "...on the surface of the moon" means? That should give a hint. Also, remember that the height, like you said, is given by

$$h(t)=h_{0}+v_{0}t+\frac{at^{2}}{2}$$

where a is the acceleration.