Maximum Height is giving me negative Hey guys for this parametric equations its giving me negative Question is:

A dart is thrown from a point 5 feet above the ground with an inital velocity of 58 ft/sec and angle of elvation of 41∘. Assume the onnly force acting on the dart is gravity. What is the maximum height reached by the dart? When and will the dart hit the ground? SHOW ALL WORK.

So what i did was
$x_t = 58 \frac{ft}{sec} * \cos(41^\circ)*t$
$y_t = -16t^2 + 58 * \sin(41^\circ)$
a When does ball land)
$-16t^2 + 58*\sin(41^\circ) * t = 0$ 
$t = 2.38sec$
b Where will it hit ground)
$x_t = 58 \frac{ft}{sec} \cos(41^\circ)*t$
$= 58 \frac{ft}{sec} \cos(41^\circ) * 2.38$
$=104.18$ ft far
All above i got correct im preety sure.. But for getting Max Height i dont get it right somehow i get negative. Im not doing it corectly?:
$y_t = -16t^2 + 58 * \sin(41^\circ)*t = 0$
$=-16(2.38^2) + 57 * \sin(41^\circ)*2.38$
And then when you solve.. it gives you negative.. Im confused here.. how i solve max height?
Remeber I am in Precalculus NOT Calculus
Equations(In my notes..)

Examples


EDIT
Ok for MAX Height what i did was take this
$y_t = -16*t^2 + 58 * \sin(41^\circ)*t=0$
When i put in calculator its fine... i get 22.055 but i remembereed the 5... So now i have to go back to finding T:
$-16t^2 + 58* \sin(41^\circ) * t + 5 = 0$
When i tried to do that.. i get wierrrrd squarerrots and stuff.. Im not sure im right
 A: I will begin where you write your two equations of interest now including the initial height information in y:
\begin{array}{l}
x_t = 58 (\text{ ft/s})\cos{(41^{\circ})} t \\
y_t = -16 (\text{ ft}/\text{s}^2) t^2 + 58 (\text{ ft/s})\sin{(41^{\circ})} t
 +5 \text{ ft}
\end{array}
To find the time to hit the ground we set the second equation equal to zero and solve
$$
y_t = -16 (\text{ ft}/\text{s}^2) t^2 + 58 (\text{ ft/s})\sin{(41^{\circ})} t
 +5 \text{ ft} = 0
$$
This is quickly solved via quadratic formula
$$
t = \frac{- 58\sin(41^{\circ}) \pm \sqrt{(58\sin(41^{\circ}))^2 - 4 (-16)(5)}}{2(-16)} \approx 2.503 \text{ s}
$$
There are two roots but typically we reject negative answers for t.
I will leave the maximum distance calculation up to you as you were doing it right but with the wrong time.
As for the max height calculation, we recognize the equation for y as a parabola. From algebra we know that the max/min of a parabola occurs at the vertex. The expression for the vertex is given by (for y):
$$
t = -\frac{b}{2a} = -\frac{58\sin(41^{\circ})}{2(-16)} \approx 1.189 \text{ s}
$$
We now take this time and plug it back into the equation for y to obtain the maximum height. I hope this helps :)
A: Make sure you have a full grasp of all the kinematics equations (this seems to be a basic physics problem). We will use this in this approach.
(a) Use $v_{f}^{2} = v_{i}^{2} + 2a\Delta y,$ where $v_{i}$ and $v_{f}$ are initial and final (vertical) velocities, respectively, $a$ is acceleration, and $\Delta y$ is the vertical displacement.
At the maximum height, the dart has $0$ vertical velocity (it is changing directions - up to down). Since gravity is the only force acting on the dart, use $a = -g = -32$ feet per second squared (up is the positive direction). Further, we know that $v_{i}$ must be $58\sin(41).$ Solving, we have
$$\Delta y = \frac{v_{f}^{2} - v_{i}^{2}}{2a}$$
$$= \frac{0 - (58\sin(41))^{2}}{2 * -32}$$
$$= 23 \text{ft}.$$
This means that the maximum height is $5 + 23 = \boxed{28 \text{ft}}.$
You seem to have figured out the other parts.
