Given :
Initial velocity, u = 12.5 m/s.
Height of camera, h = 64.3 m.
Acceleration due to gravity, g = 9.8 m/s².
To Find :
How long does it take the camera to reach the ground.
Solution :
By equation of motion :
Putting all given values, we get :
t = 2.56 and t = −5.116.
Since, time cannot be negative.
t = 2.56 s.
Therefore, time taken is 2.56 s.
Hence, this is the required solution.
Answer:
The particle’s velocity is -16.9 m/s.
Explanation:
Given that,
Initial velocity of particle in negative x direction= 4.91 m/s
Time = 12.9 s
Final velocity of particle in positive x direction= 7.12 m/s
Before 12.4 sec,
Velocity of particle in negative x direction= 5.32 m/s
We need to calculate the acceleration
Using equation of motion
Where, v = final velocity
u = initial velocity
t = time
Put the value into the equation
We need to calculate the initial speed of the particle
Using equation of motion again
Put the value into the formula
Hence, The particle’s velocity is -16.9 m/s.
Explanation:
The force of attraction due to Newton's gravitation law is
F =
Here G is the gravitational constant
m₁ is the mass of one student
m₂ is the mass of second student .
and r is the distance between them
Thus r =
If we substitute the values in the above equation
r =
= 2.46 m