Help please i need to solve this with derivatives
A ball is thrown vertically upward from a height of 6 ft with an initial velocity of 60 ft per second. How high will the ball go?

Question

Help please i need to solve this with derivatives
A ball is thrown vertically upward from a height of 6 ft with an initial velocity of 60 ft per second. How high will the ball go?

anonymous · Answer

174.75

zehanz · Answer

General formula for anobject with a stating speed $v_0$ formula  that is subjected to an accelleration a is: $v_t=v_0+at$. Use g instead of a (it is a negative number, because it tries to slow the object down) and  solve $v_t=0$ for t.

anonymous · Answer

those formulas are also with derivatives?

zehanz · Answer

Then put the result in the formula for the distance covered during accellerated movement:
$S_t=v_0t=\dfrac{1}{2}at^2$, again using g instead of a (and with a negative sign)

anonymous · Answer

I have to solve it with calculus and derivatives though, thats physics right?

zehanz · Answer

Typo: $S_t=v_0t+\frac{1}{2}at^2$
As you can see, the derivative of this formula is the first formula.

zehanz · Answer

YOu don't have to use calcuus, if you just use the formulas. Then it is algebra only.

anonymous · Answer

the thing is that i have to solve it with calculus since its for my calculus class

zehanz · Answer

OK, you can do that as follows:

First calculate how long it takes to get to zero speed:

$v_t=v_0+at=60-32.185t=0$.

Call the solution $t_1$.

Then the distance covered is the integral of the speed:

$$S=\int\limits_{0}^{t_1}v(t)dt=\int\limits_{0}^{t_1}(60-32.185t)dt$$

zehanz · Answer

32.185 being the accelleration due to gravity, in ft/s²

anonymous · Answer

ouuuu okokkk

zehanz · Answer

YW!

BTW, if you do the same with the formula $v(t)=v(0)t+at$, you get:
$$S(t)=\int\limits_{0}^{t}v(0)+a \tau d \tau=v(0)t+\frac{1}{2}at^2$$, that is the formula I suggested you at the start.