A) Let's start writing the laws of motion on both x (horizontal) and y (vertical) axis. On the x-axis, it's a uniform motion with constant velocity, while on the y-axis it's an uniformly accelerated motion:
where h=654 m is the initial height of the projectile, and are the initial velocities on the x- and y-axis, g=9.81 m/s^2 and t is the time. ad have a negative signs because they both point downward. We can calculate the initial velocity on the y-axis by requiring that , since we know that after 6.70 s the projectile reached the ground. Therefore:
from which we find
Then we can find the magnitude of the initial velocity v using the angle with respect to the vertical :
B) Let's calculate the component of the initial velocity on the x-axis:
And then, we can find how far the projectile traveled horizontally by calculating Sx at t=6.70 s, when it hits the ground:
C) The horizontal component of the velocity does not change during the motion, since it's an uniform motion on the x-axis. Therefore, vx at t=6.7 s is the same as its initial value:
D) Instead, vy changes during the motion since it's an accelerated motion, following the law
Using t=6.7 s, we can find the vertical velocity just before the projectile hits the ground
Given that a roller coaster starts from rest at A, rolls down the track to B, describes a circular loop of 12-m diameter, and travels up and down past point E. Determine the range of values of h for which the roller coaster will not leave the track at D or E. Assume no energy loss due to friction
At point A
The maximum potential energy = maximum K.E
At point A, the total energy = maximum P.E.
Down the track to point B, the P.E will be converted to maximum K.E.
Mgh = 1/2mv^2
Also, the total energy at the roller coaster will be P.E + K.E
I.e mg2r + 1/2mv^2
Where 2r = height of the loop = diameter of the loop.
Since the energy is always conserved, hence
Mgh = mg2r + 1/2mv^2
Let also consider the centripetal acceleration to keep the object in the circle.
F = mV^2 / r = mg
Mass will cancel out
U^2 = rg
Substitute that in the last equation
Mgh = mg2r + 1/2mgr
mgh = mg ( 2r + 1/2r )
Mg will cancel out
h = 2.5r
Where r = 12/2 = 6
h = 2.5 × 6
h = 15m
Therefore, the values of h for which the roller coaster will not leave the track at D or E is 15m.