Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**bella****Guest**

I was wondering if i can get some help solving this problem.. If a juggler can toss a ball into the air at a velocity of 64ft/sec from a height of 6 ft, then what is the max. height reached by the ball?

**gnitsuk****Member**- Registered: 2006-02-09
- Posts: 118

You need to use two formula:

**v = u + at**

and

**s = ut + at^2/2**

Rearrange the first one to give:

**t = (v - u) / a**

Now convert your value of 64 ft/sec into metres per second and use this as the value of u in the formula.

a = -10 (gravity acting downwards gives -'ve sign, a is not exactly 10 but I'll use 10 here). Finally v = 0 (the final velocity of the ball at the top of the motion is zero).

Put these values into the formula to get a value for t (the time to get to maximium height) now use the second foumla to find a value for s (the distance travelled). Finally add 6ft to this.

That's it.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Are you in Calculus, bella? If so, then you just start out with:

a = -32ft/sec

Where a is acceleration. Since gravity is always working on the ball, is constant, and is negative because it's "pushing" the ball downward, the acceleration is always -32.

Now we take this value and integrate it to get:

v = -32t + C. Since v at t=0 is 64,

64 = -32(0) + C, and C = 64

so v = -32t + 64

Integrate again, and we find that:

s(distance) = -16t^2 + 64t + C

But we know that the distance at time 0 is 0:

0 = -16(0)^2 + 64(0) + C, C = 0

So s = -16t^2 + 64t

And we have just derived the equations that gnitsuk gave.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline