Question

A projectile is fried from the origin (at y=0 m) as shown in the diagram. The initial velocity components are Vox=310 m/s and Voy=26 m/s. The projectile reaches maximum height at point p, then it falls and strikes the ground at point q, which is 20 m below the launch point. For how long is the projectile in the air?

3.4 s
6.0 s
6.8 s
4.5 s
9.3 s

Answer 1

3.4 s is the correct answer to the given question .

Step-by-step explanation:

As Given in the question

Vox=310 m/s

Voy=26 m/s.

As we know that

`v_(y) ^(2)= v_(ay) ^(2)\ +2gh`--------------eq(1)

As the velocity at the maximum height =0

Putting the value of
`v_(y) ^(2) \ = \ 0` in the equation(1) we get

`v_(ay) ^(2)\ +\ 2gh\ = 0\\ h=( -v_(ay) ^(2))/(2g)`

Putting the value
`v_(ay) ^(2)\` in the previous equation we get

`(\ (-26 m/s\ )^(2) )/(\ (2*\ -9.8\ ))\\ =(676)/(19.6)\\ =34.4\ m/s\\=3.4 s`

Therefore h=3.4 s