.
A marble leaves a 1.96 m high refrigerator with an initial horizontal velocity of 1.5 m/s.
(a) What is the time required for the marble to hit the ground?
(b) What is the horizontal distance between the edge of the refrigerator and the point where the marble lands?
(c) What is the velocity of the marble just before it strikes the ground?
~~~~~~~~~~~~~~~~~~~~
I will solve the problem assuming that it came from Physics.
In this problem, the marble participates in two movements, at the same time.
Vertically, the marble falls with uniform acceleration g (free fall acceleration, or gravity acceleration).
Horizontally, it moves uniformly with the constant horizontal velocity.
(a) In vertical direction, we have free fall, so
H = ,
where H is the height in meters, g = 9.81 m/s^ is the gravity acceleration, t is the time of falling.
From this formula,
t = = = 0.63 seconds (rounded).
It is the answer to the first question.
(b) In horizontal direction, the travel is UNIFORM. THEREFORE, the horizontal traveled distance is
= 1.5 * 0.63 = 0.945 meters.
It is the answer to the second question.
(c) At this part, I believe, the question is about the magnitude of the velocity,
without looking in its direction.
Then you can apply the mechanical energy conservation law
= + mgH.
Here m is the mass of the marble, H is the height.
You may cancel the common factot "m" in both sides
= + gH.
You will get then
= = = 6.38 m/s.
Solved.