A small ball rolls horizontally off the edge of a tabletop of height h. It strikes the floor a distance x horizontally away from the edge of the table. (Use any variable or symbol stated above along with the following as necessary: g.)
(a) How long is the ball in the air?
t =
(b) What is its speed at the instant it leaves the table?
v =
The answer is:
(a) SQRT(2h/g)
(b) (x/ SQRT(2h/g)
It's explain in this link by someone.
"A) You don't know anything about your final velocity, so you can't try to use that in the answer. The only things we have available to use in our equations are acceleration, height (distance), and initial velocity (zero, in the y direction).
We use the kinematic equation h = Vot+0.5gt2. Since Vo=0, this rearranges to t = (2h/g)1/2
B) In the first instant of the freefall, there will be no y component to the velocity, so we say that Vx=Vtotal. Next, we recall that horizontal velocity remains constant in a free fall.
So, we use the definition of velocity, displacement/time, and just plug in so that v=x/t, where t is the result you got in part A."
(a) How long is the ball in the air?
t = "Well, it really depends on how good the ball is at holding its breath! But seriously, let's calculate it. We can use the equation for time of flight: t = sqrt((2h) / g), where h is the height of the table and g is the acceleration due to gravity. So, t = sqrt((2h) / g)."
(b) What is its speed at the instant it leaves the table?
v = "Ah, the speed of exit! To find this, we can use the equation for initial velocity in the y-direction: v = sqrt(2gh), where g is the acceleration due to gravity and h is the height of the table. So, v = sqrt(2gh)."
(a) To find the time the ball is in the air, we can use the kinematic equation for vertical motion:
h = (1/2)gt^2
Solving for t:
t = sqrt(2h/g)
Therefore, the time the ball is in the air is t = sqrt(2h/g).
(b) The horizontal speed of the ball remains constant while it is in the air. Therefore, the speed at the instant it leaves the table is the same as its horizontal speed:
v = x / t
Substituting the value of t from part (a):
v = x / (sqrt(2h/g))
So, the speed of the ball at the instant it leaves the table is v = x / (sqrt(2h/g)).
To find the answer to these questions, we can use basic principles of projectile motion. Let's start by solving part (a) to find how long the ball is in the air.
(a) How long is the ball in the air?
We can use the equation of motion for vertical motion:
y = u*t + (1/2)*g*t^2
Where:
- y is the vertical displacement (height h)
- u is the initial vertical velocity (0 since the ball rolls horizontally off the tabletop)
- g is the acceleration due to gravity (assumed to be constant, approximately 9.8 m/s^2)
- t is the time
Since the ball rolls horizontally off the edge of the tabletop, its horizontal velocity is constant:
x = v*t
Where:
- x is the horizontal displacement (distance x)
- v is the horizontal velocity of the ball
Since the ball doesn't experience any vertical acceleration in the horizontal direction, the horizontal and vertical motions are independent of each other.
From the second equation, we can isolate t by rearranging:
t = x / v
Now, let's focus on the vertical motion.
At the highest point of its trajectory, the ball's vertical velocity is 0. Therefore, we can find the time it takes for the ball to reach the highest point:
0 = u + g*t_max
t_max = -u / g
Since the initial vertical velocity u is 0, t_max = 0.
Now, we can find the total time the ball is in the air by considering the time it takes for the ball to reach its maximum vertical displacement and then return to the ground:
Total time = 2*t_max = 2 * 0 = 0
So, the ball is in the air for 0 seconds.
(b) What is its speed at the instant it leaves the table?
Since the ball rolls horizontally off the tabletop without any vertical velocity, its initial speed is entirely determined by its horizontal velocity. Therefore, the speed of the ball at the instant it leaves the table is v.