A projectile is shot from the ground at a velocity of #5 m/s# and at an angle of #(2pi)/3#. How long will it take for the projectile to land?

Question

A projectile is shot from the ground at a velocity of #5   m/s# and at an angle of #(2pi)/3#.  How long will it take for the projectile to land?

Nathan Ashcroft · Accepted Answer

#t~~0.88s#

Assuming the launch and landing points are at equal altitudes, we can use a kinematic equation to determine the flight time from the launch of the projectile to its maximum altitude (#v_f=0#) and multiply by two to get the total flight time.

#v_f=v_i+a_yDeltat#

#=> Deltat=(v_f-v_i)/a_y#

#=>Deltat=(-v_i)/a_y#

We know that when an object is in free fall, the acceleration is equal to #-9.8m/s^2# (vertically).

Because the projectile is launched at an angle, we will need to break the velocity up into components. This can be done using basic trigonometry.

Adrian Ayala · Answer

undefined To find the time it takes for the projectile to land, you can use the formula for the time of flight of a projectile:
$$ t = \frac{2 \times v_0 \times \sin(\theta)}{g} $$
where $ v_0 $ is the initial velocity (5 m/s), $ \theta $ is the launch angle ((2π)/3), and $ g $ is the acceleration due to gravity (9.8 m/s²).

Substituting the given values:
$$ t = \frac{2 \times 5 \times \sin\left(\frac{2\pi}{3}\right)}{9.8} $$
$$ t \approx 1.02 \text{ seconds} $$