# A projectile is shot from the ground at a velocity of 5 m/s and at an angle of (7pi)/12. How long will it take for the projectile to land?

Dec 25, 2015

$0 , 9856 s$

#### Explanation:

The only force acting on the projectile after being fired is the force of gravity.

We may resolve the motion into 2 components :

In the horizontal (X) direction, there is constant velocity of $5 \cos \frac{7 \pi}{12} r a d = 5 \cos {105}^{\circ} = 5 \cos {75}^{\circ} 1 , 2941 m / 2$

Note that we take the angle with the horizontal axis as the reference angle and so the projectile is actually being fired backwards with a projection angle of ${75}^{\circ}$.

In the vertical (Y) direction, there is constant acceleration due to gravity and hence the equations of motion for uniform acceleration in 1 direction are valid.

The initial velocity in the y-direction is hence $5 \sin {75}^{\circ} = 4 , 8296 m / s$.

Therefore, time taken to reach maximum height is given by :
$v = u + a t \implies t = \frac{v - u}{a}$
$\therefore t = \frac{0 - 4 , 8296}{- 9 , 8} = 0 , 4928 s$.

Therefore the total time taken to reach ground again, since the path is symmetric, is $0 , 4928 \times 2 = 0 , 9856 s$.