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.

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step-by-step explanation:

sorry no step by step

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°