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

Answer:

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 ,

$R=v^2Sin$(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.