Question

If an object is launched at an angle of 28° with an initial velocity of 133 ft./s from a height of 6 feet how many seconds will the object hit the ground?

Answer 1

Hence after 3.98 sec i.e 4 sec Object will hit the ground .

Explanation:

Given:

Height= 6 feet

Angle =28 degrees.

V=133 ft/sec

To Find:

Time in seconds after which it will hit the ground?

Solution:

This problem is related to projectile motion for object

First calculate the Range for object and it is given by ,

`R=v^2Sin`(2Ф)/
`g`

Here R= range g= acceleration due to gravity =9.8 m/sec^2

1m =3.2 feet

So 9.8 m, equals to 9.8 *3.2=31.36 ft

So g=31.36 ft/sec^2. and 2Ф=2(28)=56

`R=133^2*Sin(56)/31.36`

`R=14664.84/31.36`

`R=467.62` fts

Now using Formula for time and range as

`R=VxT`

Vx is horizontal velocity

`Vx=V*cos`Ф

`Vx=133*cos`(28)

`Vx=117.43` ft/sec

So above equation becomes as ,

`467.62=117.43*T`

`T=467.62/117.43`

`T=3.98 sec`

T is approximately equals to 4 sec.