Physics problem dealing with projectile motion 10 pts best answer?

I'll show you both ways to do this.

θ = 45

x = 98 m

v = ?

g = -9.8 m/s²

g is negative because it's moving against gravity, causing it to decelerate vertically

First use soh cah toa to break the velocity into its x and y components, where v is the hypotenuse

Sinθ = opp / v              Cosθ = adj / v

vSinθ = opp                  vCosθ = adj

vSinθ = vᵧ                      vCosθ = vᵪ

The time in the air is

vᵪ = x / t

vCos = x / t

t = x / vCosθ

FIRST WAY:

y = vᵧt + ½gt²

0 = (vSinθ)t + ½gt²                                        when it lands, its height is y = 0

0 = (vSinθ)[x/vCosθ] + ½g[x/vCosθ]²        and its time is t = x / vCosθ

the physics part is over, now use algebra to simplify and solve for v

0 = x(Sinθ/Cosθ) + ½g[x²/v²Cos²θ]

0 = xTanθ + gx² / 2v²Cos²θ                           trig identity: Sinθ/Cosθ = Tanθ

-gx² / 2v²Cos²θ = xTanθ

-gx² / xTanθ = 2v²Cos²θ

-gx² / 2Cos²θ(xTanθ) = v²

-(-9.8)(98²) / 2Cos²(45)(98Tan(45)) = v²

960.4 = v²

30.99 m/s = v

SECOND WAY:

y = vᵧt + ½gt²

0 = (vSinθ)t + ½gt²            you have a quadratic

0 = t(vSinθ + ½gt)             to solve for t, factor out a t and set each factor equal to 0

t = 0                              vSinθ + ½gt = 0

time when hit             vSinθ = -½gt

                                       2vSinθ / -g = t

                                       time when landed (total time in the air)

But we know the total time in the air is also t = x / vCosθ

2vSinθ / -g = x / vCosθ

(2vSinθ)(vCosθ) = -gx

2v²SinθCosθ = -gx

v² = -gx / 2SinθCosθ

v² = -(-9.8)(98) / 2Sin(45)Cos(45)

v² = 960.4

v = 30.99 m/s

tsk tsk tsk! Vertical velocity my dear, but I won't answer this for you.