A missile is launched from the ground with a speed of 100 m/s at an initial angle of 45 degrees to the ground. In what direction is it moving just 2 seconds before hitting the ground?

Question

anonymous · Answer

calculate the total time first:

In the y-direction: Dy = 0 m, V1y = 70.71 m/s, a = 9.8 m/s^2.

Take one of the kinematic equations to find time,

$$dy = v1yt + 1/2at^2$$

$$0 = 70.71t - 4.9t^2$$

use quadratic formula to solve for t,

t = 14.43 s

14.43 s - 2 s= 12.43 s (because the question says 2 seconds before hitting the ground)


Therefore, we have Dy = 0 m, V1y = 70.71 m/s, a = -9.8 m/s^2, t = 12.43

Now, find the vertical component of the final velocity.

Vf = V2y + V2x


Take one of the kinematic equations again to find V2y.

$$v2y = v1y + at$$

$$v2y = 70.71 - 9.8(12.43)$$

V2y = -51.104 m/s (the negative sign indicates the velocity in the down direction)


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Find the V2x component.

V2x = 100cos45 = 70.71 m/s

Use the tangent ratio to find the angle.


$$\tan \theta = opp/adj$$

$$\tan \theta = 51.104/70.71$$

$$\tan^{-1} (51.104/70.71) = 35.8 $$

Therefore, the missile is moving at a direction of 35.8 degrees below the horizontal 2 seconds before hitting the ground.