Answer: 7.66 m/s
Explanation:
This situation is related to free fall (vertical motion). Hence, this can be considered a one-dimension problem and we can use the following equation:
Where:
is the final velocity of the egg at the asked height
is the initial velocity of the egg
is the acceleration due gravity
is the distance at which the egg is from the nest, when it is 
from the ground
Isolating :
Substituting the known values:
This is the final velocity of the egg
The
maximum altitude that the bullet will reach is the point at which its velocity
is zero. The equation that may be used in order to determine the altitude is,
<span> D = ((Vi)2 –
(Vf)2)/2g</span>
Where
Vi and Vf are the initial and final velocities, respectively. g is the
deceleration due to gravity.
Substituting
the known values,
<span> D
= ((700 m/s)2 – (0 m/s)2) / (2(-9.8 m/s2))</span>
<span> D = 25000 m</span>
Thus,
the maximum height is 25000 m.
For
the time needed to reach it, we use the equation,
<span> Vf
= Vi – (g)(t)</span>
Substituting,
<span> 0 = 700 m/s
– (-9.8 m/s2)(t)</span>
The
value of t is equal to 71.43 s.
<span> </span>
Answer:
Canyon is 50.176 meter deep.
Explanation:
The students is standing on the rim of the canyon and drops down a rock from the rim(cliff). We have to find what is the depth of the canyon i.e. how much below is the ground from the cliff.
Given data:
Time = t = 3.2 s
Initial velocity = 
= 0 m/s
Gravitational acceleration = g = 9.8 m/s²
Height = h = ?
According to second equation of motion
As initial velocity is zero, So the first term of right hand side of above equation equal to zero
h = (0.5)(9.8)(3.2)²
h = 50.176 m
This means, the rock traveled 50.176 meters to reach the bottom of the Canyon. So, the Canyon is 50.176 meter deep.
Answer:
The load has a mass of 2636.8 kg
Explanation:
Step 1 : Data given
Mass of the truck = 7100 kg
Angle = 15°
velocity = 15m/s
Acceleration = 1.5 m/s²
Mass of truck = m1 kg
Mass of load = m2 kg
Thrust from engine = T
Step 2:
⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:
T = (m1+m2)*g*sinθ
⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes m1*gsinθ .
Resultant force on truck is F = T – m1*gsinθ
F causes the acceleration of the truck: F= m*a
This gives the equation:
T – m1*gsinθ = m1*a
T = m1(a + gsinθ)
Combining both equations gives:
(m1+m2)*g*sinθ = m1*(a + gsinθ)
m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ
m2*g*sinθ = m1*a
Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:
m2*g*sinθ = (7100 – m2)*a
m2*g*sinθ = 7100a – m2a
m2*gsinθ + m2*a = 7100a
m2* (gsinθ + a) = 7100a
m2 = 7100a/(gsinθ + a)
m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)
m2 = 2636.8 kg
The load has a mass of 2636.8 kg
