Question

A 0.52 kg projectile is fired into the air from the top of a 6.31 m cliff above a valley. Its initial velocity is 9.81 m/s at 39◦ above the horizontal. Cliff x 39 9 ◦ .81 m/s 6.31 m How long is the projectile in the air? The acceleration due to gravity is 9.8 m/s 2 . Answer in units of s.

Answer 1

The projectile flies 1.93 seconds in the air.

Step-by-step explanation:

Consider the vertical motion of projectile,

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 9.81 x sin 39 = 6.17 m/s

Acceleration, a = -9.81 m/s²

Displacement is zero when the shell reaches back to ground, s = -6.31 m

Time, t =?

Substituting,

s = ut + 0.5 at²

-6.31 = 6.17 x t + 0.5 x (-9.81) x t²

4.905 t² - 6.17 t - 6.31 = 0

`t=(-(-6.17)\pm √((-6.17)^2-4* 4.905* (-6.31)))/(2* 4.905)\\\\t=(6.17\pm √(161.87))/(9.81)\\\\t=(6.17\pm 12.72)/(9.81)\\\\t=1.93s\texttt{ or }t=-0.67s`

Negative time is not possible,

The projectile flies 1.93 seconds in the air.