Question

A stone is thrown outward from the top of a 59.4-m high cliff with an upward velocity component of 19.5 m/s. How long is stone in the air?

Answer 1

Answer:

There are 2 cases here:

a. The stone travels to another certain height until its velocity becomes 0.

i.e V² = u² - 2gh (given u = 19.5m/s)

When it reaches maximum height V = 0  

so, h = u²/2g = (19.5)²/(2*9.8) => h= 19.4m

b. Now the body starts to fall freely under gravity, so:

h = 1/2*gt²

We know maximum height reached by the body = height of building + height traveled after throw in (a)

or., t = √(2h/g)

or, t = √2*(19.4+59.4) /9.8 = 4 second

Explanation:

Answer 2

Answer:

t = √2*(19.4+59.4) /9.8 = 4 sec

Explanation:

Case 1 :

The stone travels to another certain height until its velocity becomes 0.

i.e V² = u² - 2gh (given u = 19.5m/s)

When it reaches maximum height V = 0

so, h = u²/2g = (19.5)²/(2*9.8) => h= 19.4m

Case 2 :

Now the body starts to fall freely under gravity, so:

h = 1/2*gt²

We know maximum height reached by the body = height of building + height traveled after throw in case 1.

or., t = √(2h/g)

or, t = √2*(19.4+59.4) /9.8 = 4 second