Question

a rifle bullet is fired vertically upwards, leaving the course with an initial velocity of 505m / s. When will the ball at a height of 500m located? (for the displacement x, x = vit + gt ^ 2/2) take g = 10m / s ^ 2
A. after 1s and after 100s
B. after 1s because after 100s is not physically possible because of the great initial speed.
C. after 100s because after 1s is not physically possible because of the great height.
D. after 1s because the solution "after 100s" is wrong because movement slows down after acceleration.

Answer 1

Δy = (v_i)t + 1/2 * (a_y)t^2 (Technically it's t - t_0 and not t but since t_0 can always be taken as 0 then we can ignore it.)

(500 m - 0 m) = (505 m/s)(t) - 1/2 gt^2
(500 m) = (505 m/s)(t) - 1/2 gt^2
(500 m) = (505 m/s)(t) - 5t^2
-5t^2 + 505t - 500 = 0
5t^2 - 505t + 500 = 0

Use the quadratic formula to get t = 1 and t = 100. This means that at 1 second and 100 seconds, the ball is at a height of 500m above the ground (we define the ground to be 0 m on the y-axis).

So, the answer is A. Though when it says "after," it means right after; for example 1.000000000000000000000000000000001 and 100.0000000000000001 seconds with infinite zeros instead of the finite amount I have to put here because it's physically impossible to put an infinite amount of zeroes.

Also, intuitively, the reason why the ball is at a height 500 metres above the ground twice is because it's going upward really fast and being slowed down by gravity and it's going fast enough to go higher than 500 metres and then it goes back down when gravity wins and reaches a height of 500 metres again except this time it's going downward instead of upward. So, the ball has been above 500 metres for 100 – 1 = 99 seconds.

Hope this helps.