Question

An antiaircraft is fired vertically upward with a muzzle of 1000m/s. Calculate the instantaneous velocities of at the ends of 20 and 50 sec. When its height will be 37.5km.

Answer 1

a) Let's take the upwards as the positive direction. Then, we can find the maximum height from the kinematic equation:

v} = v? + 2ah,

here, Vf = 0 ms -1 is the final velocity of the shell at the maximum height, vi is the initial velocity of the shell, a = g = -9.8 ms -2 is the acceleration due to gravity, h is the height.

Then, we get:

0 = (488 ms -1)2 + 2.(-9.8 ms-2).h, 19.6 m s2. h = 238144 m?s-2,

238144 ms-2

h = !! 12.15. 103 m = 12.15 km.

19.6 ms-2

b) We can find the time that shell takes to reach the maximum height from the kinematic equation:

Uf = Vi + at,

here, Vf = = 0 ms-1 is the final velocity of the shell at the maximum height, v; is the initial velocity of the shell, a = g = -9.8 ms -2 is the acceleration due to gravity, t is the time.

Then, we get:

0 = 488 ms -1 +(-9.8 ms-2).t, 9.8 ms-2.t = 488 ms-1,

t 9.8 ms-2 = 49.8 s.

c) We can find the instantaneous velocity at the end of 40 s from the kinematic

equation:

Vp = Vi + at = 488 ms -1 + (-9.8 ms-2). 40 s = 96 ms -1.

d) Similarly, we can find the instantaneous velocity at the end of 60 s:

Vp = vi + at = 488 ms -1 + (-9.8 ms-2).60 s = -100 ms -1.

The sign minus indicates that the velocity of the shell is directed downward (the shell

is begin to fall).

Answer:

a) h = 12.15 km

b) t 49.8 s

c) vf = 96 ms-1

d) vf = -100 ms-1

hope it helps

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