A bullet fired vertically upwards reaches a height of 500m. [g=9.8m/s2]. What is its initial velocity?
0=v0^2 +2(-9.8m/s^2)(500m)
0=u^2-19.6+1000
0=u+980.4
0-980.4=v0^2
V0^2=-980.4m/s
V^2 = Vo^2 + 2g*h.
V = 0, g = -9.8 m/s^2, h = 500 m.
Vo = ?.
@Nataizya, why did you add 1000
-->0=u^2-19.6+1000
instead of multiplying?
--->0=u^2-(19.6*1000)
To find the initial velocity of the bullet, we can use the kinematic equation for motion in the vertical direction:
vf^2 = vi^2 + 2aΔy
Where:
vf - final velocity (0 m/s when reaching maximum height)
vi - initial velocity (what we need to find)
a - acceleration due to gravity (9.8 m/s^2)
Δy - change in position (500 m upwards)
Since the bullet reaches its maximum height and then falls back down, we know that its final velocity is zero (vf = 0 m/s). Plugging in the known values into the equation, we have:
0 = vi^2 + 2(9.8 m/s^2)(500 m)
Simplifying the equation:
0 = vi^2 + 9800 m^2/s^2
Rearranging the equation to solve for vi:
vi^2 = -9800 m^2/s^2
Taking the square root of both sides:
vi = √(-9800 m^2/s^2)
The square root of a negative value is imaginary, which means there is no real solution for the initial velocity of the bullet. Therefore, we cannot determine the initial velocity with just the given information.