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

Answer:

3.4 s is the correct answer to the given question .

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

$v_{ay} ^{2}\ +\ 2gh\ = 0\\ h=\frac{ -v_{ay} ^{2}}{2g}$

Putting the value $v_{ay} ^{2}$ in the previous equation we get

$\frac{\ (-26 m/s\ )^{2} }{\ (2*\ -9.8\ )}\\ =\frac{676}{19.6}\\ =34.4\ m/s\\=3.4 s$

Therefore h=3.4 s