If a stone is thrown vertically upward from the surface of the moon with a velocity of 10 m/s, its height (in meters) after t seconds is h(t)=10t−0.83t^2.
(a) What is the velocity of the stone after 8 seconds?
(b) What is the velocity of the stone after it has risen 25 m?

Question

If a stone is thrown vertically upward from the surface of the moon with a velocity of 10 m/s, its height (in meters) after t seconds is h(t)=10t−0.83t^2. 
(a) What is the velocity of the stone after 8 seconds? 
(b) What is the velocity of the stone after it has risen 25 m?

anonymous · Answer

I already figured out part a. I just need assistance with b! Please help!

anonymous · Answer

Omg. It's supposed to say h(t)=10t-0.83t^2.

badhi · Answer

since the equations on constant acceleration stands,

your given expression is similar to $$S=ut +\frac{1}{2}at^2$$
here $$u=10,\quad a=0.83	imes 2=1.66$$

to find the velocity at 25m use $$v^2=u^2+2as$$