Question

A particle 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.

Answer 1

Answer:

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:

• Velocity in the Y axis:

Voy=Vo*sinα

• Time to reach the maximal height :

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 ))

• Time required to return to the starting point:

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 ))

• α = 0º , sinα=0

The particle goes totally horizontal, goes away indefinitely

T total= 2*Voy/(g*sin( α )) ≅∞

• α = 90º, sinα=1

T total=2*Voy/g