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:
SKID
Explanation:
In general, airplane tracks are flat, they do not have cant, consequently the friction force is what keeps the bicycle in the circle.
Let's use Newton's second law, let's set a reference frame with the horizontal x-axis and the vertical y-axis.
Y axis y
N- W = 0
N = W
X axis (radial)
fr = m a
the acceleration in the curve is centripetal
a =
the friction force has the expression
fr = μ N
we substitute
μ mg = m v²/r
v =
we calculate
v =
v = 1,715 m / s
to compare with the cyclist's speed let's reduce to the SI system
v₀ = 18 km / h (1000 m / 1 km) (1 h / 3600 s) = 5 m / s
We can see that the speed that the cyclist is carrying is greater than the speed that the curve can take, therefore the cyclist will SKID