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?

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

Step-by-step 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 ,

(2Ф)/

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

 fts

Now using Formula for time and range as

Vx is horizontal velocity

Ф

(28)

 ft/sec

So above equation becomes as ,

T is approximately equals to 4 sec.

2 and 1.

step-by-step explanation:

plug in each answer. 0 doesn't work, 1 does. so does 2. 4 doesn't:

0 = 1 is not true.

2 = 2 is true.

4 = 4 is true.

8 = 16 is not true.

hope this !

f(x) is going down by 4

step-by-step explanation:

(its either that or its going left by 4)