Thread: damsel in distress- need help with this questions

1. damsel in distress- need help with this questions

1.Assuming a constant rate of acceleration due to gravity of 32 feet per second per
second, findnd the initial velocity needed to launch a projectile from the surface of the
earth to a height of 900 feet.

2. Suppose that the probability distribution function for customers waiting to be
seated at a certain restaurant on a certain evening has the form f(t) = ce^-ct for
t > 0, where c = 1/10 is the reciprocal of the average waiting time in minutes. Find
the time T such that exactly 5% of customers wait longer than T minutes to be
seated. Round your answer to the nearest minute.

2. Re: damsel in distress- need help with this questions

No.1 You have to learn the formulae for these questions
The one you need is v^2=u^2+2as where u = initial speed v=final speed a=acceleration and s= distance
For you v=0 s=900 a=-32 (because it is traveling upwards) and u is the question
So 0=u^2-(2*32*900) u^2=57600 u= 240

3. Re: damsel in distress- need help with this questions

You could also do this problem using the fact that if an object has constant acceleration of -32 f/s^2, its velocity, after t seconds, is -32t+ v0, where v0 is the speed at t= 0, and its height is -16t^2+ v0t+ h0, where h0 is the height when t= 0. You can take h0= 0 and you want to find v0 so that the maximum height (this is a quadratic so the maximum height is at the vertex of the parabola- you can find it by completing the square) is 900.

For the second problem, you want T such that $\int_T^\infty f(t)dt= 0.05$.

4. Re: damsel in distress- need help with this questions

Hello, iva17!

1. Assuming a constant rate of acceleration due to gravity of 32 ft/sec2,
find the initial velocity needed to launch a projectile from the surface of the earth to a height of 900 feet.

The height $y$ of the projectile is given by: . $y \;=\;h_o + v_ot - \tfrac{1}{2}gt^2$

. . where: . $\begin{Bmatrix}h_o &=& \text{initial height} \\ v_o &=& \text{initial velocity} \\ g &=& \text{accel'n due to gravity} \end{Bmatrix}$

We are given: . $\begin{Bmatrix}h_o &=& 0 \\ g &=& 32\end{Bmatrix}$

The formula becomes: . $y \;=\; v_ot - 16t^2$

Assuming we want the minimum initial velocity to reach a height of 900 feet,
. . the vertex of the parabola is: . $t\;=\;\frac{\text{-}b}{2a} \:=\:\frac{\text{-}v_o}{2(\text{-}16)} \:=\:\frac{v_o}{32}$

The formula becomes: . $v_o\left(\frac{v_o}{32}\right) - 16\left(\frac{v_o}{32}\right)^2 \;=\;900 \quad\Rightarrow\quad \frac{v_o^2}{64} \:=\:900$

. . . . . $v_o^2 \:=\:64\cdot900 \quad\Rightarrow\quad v_o \:=\:8\cdot30$

Therefore: . $v_o \:=\: 240\text{ ft/sec}$