1. ## Projectile problem

Problem requires the following to be answered: 1) Range, 2) total flight time and 3) max altitude. I need someone to review my range answer and my total flight time answer. In addition, I need some help setting up my max altitude.

The equation is (vcosθ)ti +[(vsinθ)t -1/2 gt^2]j
To get range, I set y=0 and solved for t.
y= (vsinθ)t -1/2 gt^2
0= (vsinθ)t -1/2 gt^2
1/2 gt^2 = (vsinθ)t
gt^2= 2(vsinθ)t
gt= 2(vsinθ)
t= 2(vsinθ)/g TOTAL FLIGHT TIME
Inserted t into x= vcosθt
vcosθ * 2(vsinθ)/g
2 sinθcosθv^2/g
Sin2θv^2/g RANGE

Need help on setting up max altitude

Problem requires the following to be answered: 1) Range, 2) total flight time and 3) max altitude. I need someone to review my range answer and my total flight time answer. In addition, I need some help setting up my max altitude.

The equation is (vcosθ)ti +[(vsinθ)t -1/2 gt^2]j
To get range, I set y=0 and solved for t.
y= (vsinθ)t -1/2 gt^2
0= (vsinθ)t -1/2 gt^2
1/2 gt^2 = (vsinθ)t
gt^2= 2(vsinθ)t
gt= 2(vsinθ)
t= 2(vsinθ)/g TOTAL FLIGHT TIME
Inserted t into x= vcosθt
vcosθ * 2(vsinθ)/g
2 sinθcosθv^2/g
Sin2θv^2/g RANGE

Need help on setting up max altitude
Yes, your formula is correct. If t is the time of flight, then $x=v_{ix}t \implies t=\frac{x}{v_{ix}}$. Consider the vertical motion alone; when the flight is over and the object strikes the target, vertical displacement $=0=v_{iy}t + \frac{1}{2} (-g)t^2$. Solving this equation gives $t=2v_{iy}/g$. But $t=x/v_{ix}$, so

$\frac{x}{v_{ix}} = \frac{2v_{iy}}{g}$ or: $x=\frac{2v_{ix}v_{iy}}{g} = \frac{2(v_i cos \theta)(v_i sin \theta)}{g}$

You can now simplify the numerator using the double angle formula

$x= \frac{v^2_{i} sin 2 \theta}{g}$

3. So both of my TOTAL FLIGHT TIME and range answers are correct???

Problem requires the following to be answered: 1) Range, 2) total flight time and 3) max altitude.
I need someone to review my range answer and my total flight time answer.
In addition, I need some help setting up my max altitude.

The equation is: . $\vec s \;=\; (v\cos\theta)\vec i + \left[(v\sin\theta)t -\tfrac{1}{2}gt^2\right]\vec j$

To get range, I set $y=0$ and solved for $t.$

. . $(v\sin\theta)t - \tfrac{1}{2}gt^2 \:=\:0 \quad\Rightarrow\quad \tfrac{1}{2}gt^2 \:=\: (v\sin\theta)t \quad\Rightarrow\quad gt^2\:=\:2(v\sin\theta)t$ . $\Rightarrow\quad gt \:=\:2(v\sin\theta)$

. . $t \:=\: \frac{2v\sin\theta}{g}$ . . . Total flight time . . Yes!

Inserted $t$ into $x\:=\v\cos\theta)t" alt="x\:=\v\cos\theta)t" />

. . $x \:=\v\cos\theta)\left(\frac{2v\sin\theta}{g}\right) \;=\; \frac{v^2(2\sin\theta\cos\theta)}{g} \;= \;\frac{v^2\sin2\theta}{g}" alt="x \:=\v\cos\theta)\left(\frac{2v\sin\theta}{g}\right) \;=\; \frac{v^2(2\sin\theta\cos\theta)}{g} \;= \;\frac{v^2\sin2\theta}{g}" /> . . . Range . . Right!

Need help on setting up max altitude.

The height function is: . $y \:=\v\sin\theta)t - \tfrac{1}{2}gt^2" alt="y \:=\v\sin\theta)t - \tfrac{1}{2}gt^2" />

Its graph is a down-opening parabola: $\cap$

Its maximum is at its vertex: . $t \;=\;\frac{-b}{2a} \;=\;\frac{-v\sin\theta}{2(-\frac{1}{2}g)} \;=\;\frac{v\sin\theta}{g}$

Insert into $y\!:\;\;(v\sin\theta)\left(\frac{v\sin\theta}{g}\r ight) - \frac{1}{2}g\left(\frac{v\sin\theta}{g}\right)^2$

Simplify: . $\frac{v^2\sin^2\theta}{g} - \frac{1}{2}\!\cdot\!\frac{v^2\sin^2\theta}{g} \;\;=\;\; \frac{v^2\sin^2\theta}{2g}$ . . . maximum altitude