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KiRa [710]
3 years ago
15

A missle is fired horizontally with an initial velocity of 45 m/s from the top of a building 75 m high.

Physics
1 answer:
NARA [144]3 years ago
4 0

The horizontal range of the missile is b) 176 m

Explanation:

The motion of the missile is a projectile motion, so it consists of two independent motions:  

- A uniform motion with constant velocity along the horizontal direction  

- A uniformly accelerated motion with constant acceleration (equal to the acceleration of gravity) in the vertical-downward direction  

To find the time of flight of the missile, we study the vertical motion. We can use the following suvat equation:

s=u_y t+\frac{1}{2}at^2

where:

s = 75 m is the vertical displacement of the missile (the height of the building)

u_y=0 is the initial vertical velocity  (the missile is thrown horizontally)

t is the time of flight

a=g=9.8 m/s^2 is the acceleration of gravity

Solving for t, we find the time of flight:

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(75)}{9.8}}=3.91 s

This means that the missile takes 3.91 s to reach the ground.

Now we study the horizontal motion: the missile moves with a constant horizontal velocity of

v_x = 45 m/s

Therefore, the distance covered in a time t is

d=v_x t

and by substituting t = 3.91 s, we find the horizontal range of the missile:

d=(45)(3.91)=176 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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<h3>Data obtained from the question</h3>
  • Initial velocity (u) = 8 m/s
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