Michael asked in Science & MathematicsPhysics · 6 months ago

# Physics help please Thank you?

Thank you so much :)

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• 6 months ago

(a)

The optimal angle is 45°

(b)

Given: v₀ = 50 m/s at 45°

Find the time to maximum height using the initial velocity.

vᵧ = v₀ᵧ – gt

0 = 50sin(45°) - (9.8 m/s²)t

t = 3.61s

Time to maximum distance is twice the time to maximum height.

s = v₀ₓt

s = 50cos(45°)(3.61)(2)

s = 255 m

• 6 months ago

Using general results the horizontal range is given by:

(v₀²/g)sin2θ

The maximum value is when sin2θ = 1

i.e. 2θ = 90°, so when θ = 45°

Hence, maximum range is v₀²/g

so, (50)²/g => 255.1 metres

:)>

• oubaas
Lv 7
6 months ago

vertical level motion :

0 = Vo*sin Θ *t -g/2*t^2

Vo*sin Θ = g/2*t

t = 2Vo*sin Θ / g

horizontal motion :

range = Vo*cos Θ*t = Vo*cos Θ*(2Vo*sin Θ / g) = Vo^2/g*2*sin Θ*cos Θ

trigonometry helps us , since 2*sin Θ*cos Θ = sin 2Θ

sin 2Θ is max and equal to 1 when 2Θ = 90° , which means Θ = 45°

max range = 50^2/g = 2500/9.8066 ≅ 255 m