a bullet is fired in the air vertically from ground level with an initial velocity of 262 m/s. find the maximum height and maximum velocity.

Question

a bullet is fired in the air vertically from ground level with an initial velocity of 262 m/s.  find the maximum height and maximum velocity.

rogue · Answer

The equation for free-fall is$$s(t) = - \frac {1}{2} g t^2 + v_0 t + s_0$$Where s(t) is the position of the object, g is the gravitational acceleration, t is time, Vo is initial velocity, So is initial height.

In your case, So = 0 m, g = 9.8 m/s^2, Vo = 262 m/s$$s(t) = -4.9 t^2 + 262t$$Its max height will be at the maximum of the parabola, which is at x = -b/2a.$$x = \frac {-b}{2a} = \frac {-262}{-9.8} = 26.735 s$$

rogue · Answer

Its maximum velocity will be at the initial time. Velocity is the derivative of position, so$$s' (t) = v(t) = -9.8t + 262$$Plugging in 0 for time, we just get that the max velocity is 262 m/s.

rogue · Answer

Oh, and to find the maximum height, plug in the 26.735 s for time in the position equation.$$s(26.735) = -4.9 (26.735)^2 + 262(26.735) = 3502 m$$

rogue · Answer

Do you get what I did or need me to explain something?