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:
<em>This problem is related to projectile motion for objec</em>t
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.
Answer:
x = ±3 sqrt(2)
Step-by-step explanation:
x^2 + x^2=6^2
Combine like terms
2x^2 = 6^2
Divide each side by 2
x^2 = * 6^2/2
Take the square root of each side
sqrt(x^2) =± sqrt( 6^2/2)
Remember that sqrt(a/b) = ±sqrt(a ) /sqrt(b)
x = ±sqrt(6^2)/sqrt(2)
x = ±6 /sqrt(2)
We do not leave square roots in the denominator, so we multiply the top and bottom by sqrt(2)
x = ±6 /sqrt(2) * sqrt(2)/ sqrt(2)
x = ±6 *sqrt(2) /( sqrt(2)*sqrt(2))
x = ±6 sqrt(2) /(2)
x = ±3 sqrt(2)
the answer should be C. 5.27
4.572 kilometers is the answer.