A projectile is shot from the ground at a velocity of #1 ms^-1# at an angle of #pi/3#. How long will it take for the projectile to land?

Question

A projectile is shot from the ground at a velocity of #1 ms^-1# at an angle of #pi/3#.  How long will it take for the projectile to land?

Isabella Alvarez · Accepted Answer

The total time taken for the projectile to return to earth will be 0.176 s.

The vertical component of the initial velocity, #v_y#, is given by #v_y = vsintheta = 1 sin(pi/3) = 0.866 ms^-1# At the top of its path the vertical component of the velocity will be #0#. Use the formula #v = u + at# where: Rearranging to make the time the subject: #t = (v-u)/a = (0-0.866)/-9.8 = 0.088 s# This is the time from launch to the top of the trajectory, and the projectile will take the same amount of time to return to earth, so this time should be doubled for the whole motion, so #t= 0.176 s#. This makes sense as an answer due to the very low initial velocity.

Nora Arzate · Answer

undefined To find the time it takes for the projectile to land, you can use the equation for the time of flight of a projectile:

$$ T = \frac{2 \times V_0 \times sin(\theta)}{g} $$

where:
- $ T $ is the time of flight,
- $ V_0 $ is the initial velocity (1 m/s in this case),
- $ \theta $ is the launch angle ($ \pi/3 $ radians in this case),
- $ g $ is the acceleration due to gravity (approximately $ 9.81 \, m/s^2 $).

Plugging in the values:

$$ T = \frac{2 \times 1 \times sin(\pi/3)}{9.81} $$

$$ T \approx \frac{2 \times 1 \times \sqrt{3}/2}{9.81} $$

$$ T \approx \frac{\sqrt{3}}{9.81} $$

$$ T \approx 0.277 \, seconds $$

So, it will take approximately 0.277 seconds for the projectile to land.