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2 years ago

1. A garelle running 14 m/s does not see the high cliff ahead. The next day buzzards are seen cireling a

Physics

1 answer:

katovenus [111]2 years ago

8 0

The vertical motion of a dropped object from height is a motion that is caused by gravity alone

The height of the cliff and the speed of the bullet are:

1. The height of the cliff is 19.6 meters

2. The speed of the bullet when it left the gun is approximately 2,245 m/s

The reason the above values are the correct values are given as follows:

The given velocity of the gazelle, vₓ = 14 m/s

Location where buzzards are seen, d = 28 m from the base of the cliff

The height of the cliff = Required

Solution:

The gazelle fell off the cliff

The horizontal distance the gazelle covered during free fall, d = 28 meters

The time it takes the gazelle to travel the horizontal distance, t = The time of free fall, and is given as follows;

$t = \dfrac{Horizontal \ distance}{Horizontal \ velocity} = \dfrac{d}{v_x}$

Therefore;

$t = \dfrac{28 \ m}{14 \ m/s} = 2 \ s$

Vertical distance covered in the time falling freely = Height of the cliff, h

The formula for (vertical) distance, h = u·t + (1/2)·g·t²

Where;

u = The initial vertical velocity = 0

t = 2 seconds from above

g = Constant for the acceleration due to gravity ≈ 9.8 m/s²

Therefore;

h = 0 × 2 + (1/2) × 9.8 m/s² × (2 s)² ≈ 19.6 m

The height of the cliff, h = 19.6 m

2. The known parameters:

The direction the bullet was shot = Horizontally

Height from which the bullet was shot, h = 1.4 meters above the ground

The horizontal distance the bullet travels before landing, d = 1.2 km

Required:

The speed with which the bullet left the gun, assuming no air resistance

Solution:

The vertical motion of the bullet is due to gravity and the initial vertical velocity, u = 0

Therefore;

h = u·t + (1/2)·g·t²

1.4 = (1/2) × 9.8 × t²

t² = 1.4/((1/2) × 9.8) = 2/7

t = √(2/7)

$Velocity = \dfrac{Distance}{Time}$

$The\ speed \ of \ the \ bullet, v = \dfrac{1,200 \ m}{\sqrt{\dfrac{2}{7} } s} \approx 2,245 \ m/s$

The speed of the bullet when it left the gun, v ≈ 2,245 m/s

Learn more about free fall motion here:

brainly.com/question/2148964

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