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# Physics problem?

I don't know how to do, thanks ### 4 Answers

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• Favorite Answer

Please remember to select a Favorite Answer from among your results. • vertical motion equation

hf-hi = Vo*sin 30°*t-g/2*t^2

(0-1)-1.5*0.5*t+4.903t^2 = 0

4.903t^2-3t/4-1 = 0

time t = (0.75+√(0.75^2+4.903*4)/9.806 = 0.535 sec

final speed V = √Vo^2+2gh = √1.5^2+19.612*1 = 4.68 m/sec (conservation  of energy)

or

final speed V = √(Vo*cos 30°)^2+(g*t-Vo*sin 30)^2

V = √1.5^2*3/4+(9.806*0.535-1.5*0.5)^2 = √1.688+4.50^2 = 4.68 m/sec

• s₀ = 1 m

u = 1.5 m/s

vertical component of velocity, uy = 1.5sin30°

θ = 30°

(a) When cork reaches the ground s = 0

s = s₀ + uy*t + ½gt²

0 = 1 + (1.5sin30)t + ½(-9.8)t²

4.9t² - 0.75t - 1 = 0

Using quadratic solution

t = {-(-0.75)±√[(-0.75)²-4(4.9)(-1)]}/(2*4.9)

t = 1.5 ±√[(-1.5)²-4(4.9)(-1)]}/(2*4.9)

t = 0.53 s

Cork reaches the ground in 0.53 s

(b) Let vertical component of velocity when cork reaches ground = vy

vy = uy + gt

vy = 1.5sin30 + (-9.8)(0.53)

vy = -4.4 (negative sign denotes cork travelling downwards)

horizontal component of initial velocity, ux = 1.5cos30, which will remain until the cork reaches the ground

resultant velocity =  √[(-4.4)² + (1.5cos30)²] = 4.6 m/s

speed of the cork when it reaches the ground is 4.6 m/s

• Anonymous
2 months ago

(a) vertical kinematics:

y = y₀ + Vy₀*t + ½at²

0 = 1 + 1.5*sin30º*t - ½*9.8*t² = 1 + 0.75t - 4.9t²

quadratic with roots at

t = -0.38 s ← ignore negative root

and t = 0.53 s

(b) Easiest is v² = u² + 2gh = 1.5² + 2*9.8*1 = 22 (m²/s²)

v = 4.7 m/s

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