Aparticle is projected with velocity v0 directly up a slope which makes an angle α with the horizontal. assume frictionless motiona)find the time required to return to the starting point. b)explicitly check your answer’s units and discuss the limiting cases of α = 0º and α = 90º.

a)T total = 2*Voy/(g*sin( α ))

b)α = 0º ,  T total≅∞ (the particle, goes away horizontally indefinitely)

α = 90º,  T total=2*Voy/g

Explanation:

Voy=Vo*sinα

Kinematics equation: Vfy=Voy-at

a=g*sinα ;  g is gravity

if Vfy=0 ⇒ t=T ; time to reach the maximal height

so:

0=Voy-g*sin( α )*T

T=Voy/(g*sin( α ))

After the object reaches its maximum height, the object descends to the starting point, the time it descends is the same as the time it rises.

So T total= 2T = 2*Voy/(g*sin( α ))

The particle goes totally horizontal, goes away indefinitely

T total= 2*Voy/(g*sin( α )) ≅∞

T total=2*Voy/g

b) phosphorus pentoxide

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