Question

an elevator of height 32m is ascending upwards with an acceleration of 6.2m/sec^2 .after a time t a coin is dropped from the top of the elevator .when will the coin strike the floor of the elevator

Answer 1

let t be the time taken by the coin to strike the floor of the elevator.

The coin was moving with the elevator at a velocity V .

Just after losing contact , it begins to descend with velocity V ( the law of inertia)

S1 = the net displacement of the coin downwards with respect to the initial position.

S1 = 1/2gt² - vt

S2 = The net displacement of the ascending  elevator with respect to position .

S2= vt + 1/2at²

Now S1 + S2 = H ( height of the elevator)

When we add S1 and S2 we get :

t = √{2h / (g - a)}

Doing the substitution we get:

√(64 / [9.8 -6.2])

√17.7777 = 4.22 seconds

Answer 2

Answer:

Explanation:

let t be the time taken by the coin to strike the floor of the elevator.

The coin was moving with the elevator at a velocity V .

Just after losing contact , it begins to descend with velocity V ( the law of inertia)

S1 = the net displacement of the coin downwards with respect to the initial position.

S1 = 1/2gt² - vt

S2 = The net displacement of the ascending elevator with respect to position .

S2= vt + 1/2at²

Now S1 + S2 = H ( height of the elevator)

When we add S1 and S2 we get :

t = √{2h / (g - a)}

Doing the substitution we get:

√(64 / [9.8 -6.2])

√17.7777 = 4.22 seconds