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;
[tex]t = \dfrac{Horizontal \ distance}{Horizontal \ velocity} = \dfrac{d}{v_x}[/tex]
Therefore;
[tex]t = \dfrac{28 \ m}{14 \ m/s} = 2 \ s[/tex]
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)
[tex]Velocity = \dfrac{Distance}{Time}[/tex]
[tex]The\ speed \ of \ the \ bullet, v = \dfrac{1,200 \ m}{\sqrt{\dfrac{2}{7} } s} \approx 2,245 \ m/s[/tex]
The speed of the bullet when it left the gun, v ≈ 2,245 m/s
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