Question

a projectile is launched horizontally with an initial speed 3.5 m/s from a height 4m. if the projectile is relaunched horizontally from the same height with initial speed 2vo, what is the time for the projectile to reach the ground?

Answer 1

The important distinction with this problem is to understand the vectors corresponding to this motion. Remember that vertical and horizontal motion have no influence on one another. To find the time taken to reach the ground, the initial horizontal velocity will play no role, so we only look at the vertical information.

Using the kinematic equation:\[d=v _{i}t+\frac{ 1 }{ 2 }at ^{2}\]We put in the relevant information, keeping in mind that it should all refer to the projectile's vertical motion. The horizontal motion will not affect the time it takes for the projectile to hit the ground, only the vertical acceleration due to gravity will.

Our variable values would be:
d = 4m
vi = 0 m/s
a = 9.81 m/s^2
t = ? s

Substituting in:
4 = (0)t + (1/2)(9.81)(t^2)
4 = 4.905 t^2
t^2 = 0.8155
t = 0.9 s

We have found that the time taken for the projectile to reach the ground is 0.9 s.

Now, considering what I had mentioned earlier, if we were to increase the initial velocity to twice the velocity we just had, it would make no difference. Horizontal motion does not affect vertical motion.

Therefore the time would also be:
t = 0.9 s

Answer 2

Thanks!!!