Physics problem dealing with projectile motion 10 pts best answer?

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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...Show more

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