1. ## Thrown Ball problem

if a ball is thrown vertically upward at 30 m/s, then its approxiamte height im meters t seconds later is given by
h(t)=30t-5t^2

1) what is the domain of h?
2) for what time limit is theball more than 45 feet above the ground?
3) Whatis themaximu height of the ball?
4) After how many seconds does the ball reach the maximun height?

2. Notice the word "help". You must show your work in order for help to be effective. You don't learn anything if someone else does your homework.

I'll get you started.

1) Do you have a definition of "Domain"? How long does the ball stay in the air?

2) h(t) = 45 should have two solutions. What are they? You should be good at solving quadratic equations.

3 & 4) You should know a think or two about parabolas. Find the vertex and maybe the axis of symmetry.

3. ok so far whati have on that is:
1)the ball stays in the air for 6 seconds right?
2) the ball is at 45 at 3 seconds andbefore and afterthat its below it.
3&4)x=-30/(2*-5)=3
and the maximum height is 45

4. Originally Posted by shamee
if a ball is thrown vertically upward at 30 m/s, then its approxiamte height im meters t seconds later is given by
h(t)=30t-5t^2

1) what is the domain of h?
2) for what time limit is theball more than 45 feet above the ground?
3) Whatis themaximu height of the ball?
4) After how many seconds does the ball reach the maximun height?

Originally Posted by shamee
ok so far whati have on that is:
1)the ball stays in the air for 6 seconds right?
2) the ball is at 45 at 3 seconds andbefore and afterthat its below it.
3&4)x=-30/(2*-5)=3
and the maximum height is 45
1) You are looking for the domain, which is all the possible values for t, not just the amount of time the ball is in the air.

2) This is best answered by considering for what times is h = 45 ft. So solve $30t - 5t^2 = 45$. Since your height function is a parabola that opens downward, we know that the ball is higher than 45 feet for the time interval between these two times.

3 and 4) Your height function is a parabola opening downward. So how do you find the vertex form for $h(t)=30t-5t^2$? The vertex will be the point where you have the maximum height.

-Dan

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### a ball is thrown vertically upward at 30 m s.find vgh at t=4sec

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