A tourist stands at the top of the Grand Canyon, holding a rock, overlooking the valley below. Find the final velocity, and disp
1 answer:
Answer:
a. Vf = 39.24 [m/s]
b. Vf = 31.24 [m/s]
c. Vf = 47.24 [m/s].
Explanation:
To solve this problem we can use the following equation of kinematics. We have to keep in mind that the gravitational acceleration acts downwards, therefore when the rock falls towards the abyss it has the same direction of the acceleration and that is why the gravitational acceleration has a positive sign in the equation.
where:
Vf = final velocity [m/s]
Vo = initial velocity = 0 (when the rock is dropped)
g = gravitational acceleration = 9.81 [m/s²]
t = time = 4 [s]
a.
b.
In this particular situation, the acceleration will be taken as negative because the gravity is pointing in the opposite direction of the movement of the rock.
The negative sign in the answer tells us that the rock no longer moves up instead it does downwards when 4 seconds have passed.
c.
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Given that,
The attached figure shows an airplane is traveling from North Africa to England.
To find,
The arrow that shows the direction gravity pulls the plane.
Solution,
The force of gravity always acts in the downward direction, The weight of an object is given by :
W = mg
Arrow 2 shows the direction of gravitational pull.
Answer:
C The launcher will fall off the platform and land D/2 to the left of the platform because the launcher is twice the mass of the ball.
Explanation:
The figure is missing: you can find it in attachment.
We can apply the law of conservation of momentum to check that the launcher will leave the platform with a speed which is half the speed of the ball. In fact, the total initial momentum is zero:
while the total final momentum is:
where
is the mass of the launcher
is the mass of the ball
is the velocity of the launcher
is the velocity of the ball
Since the total momentum must be conserved, , so
Therefore we find
which means that the launcher leaves the platform with a velocity which is half that of the ball, and in the opposite direction (to the left).
Since the distance covered by both the ball and the launcher only depends on their horizontal velocity, this also means that the launcher will cover half the distance covered by the ball before reaching the ground: therefore, since the ball covers a distance of D, the launcher will cover a distance of D/2.
