A ball rolls horizontally off the edge of a tabletop that is 1.80 m high. It strikes the floor at a point 1.42 m horizontally away from the table edge. (Neglect air resistance.)
(a) How long was the ball in the air?
(b) What was its speed at the instant it left the table?
i have the equation
x-x(0)= V(0)xt + 1/2at^2
y-y(0) = V(0)yt - 1/2at^2
but i dunno if i am using the right equation....can someone help me plz?
If x is horizontal, no you are not using the right equation.
Gravity has no horizontal component
You horizontal speed, call it U is constant
so
X - Xo = U t
In the vertical direction, fine with a = -g = -9.8 m/s
what is U...meaning acceleration?....
so basically i would have
1.80= -9.8t?...is that right?
Now to continue
call the initial X0 = 0
the final X = 1.42 m
so
1.42 = U t
where U is that initial speed which remains the horizontal speed and t is time in the air
NOW vertical direction
call y upwards from the top of the table
so Yo = 0
Yfinal = - 1.90 m , the floor
Vo = 0, no vertical speed originally, only the horizontal speed U
so
-1.90 = 0 - (1/2)(9.8) t^2
t^2 = 1.9/4.9 = /0.388
so
t = 0.623 seconds in the air
now back to get U
1.42 = U t = U (.623)
so
U = 2.28 m/s
whoops, I used 1.90 instead of 1.80 for table height. You will have to redo the arithmetic :)
I called U the horizonal speed, and V the vertical speed.
The point is that this is two problems.
1. a constant speed horizontal problem where distance = horizontal rate * time
2. a vertical problem where distance is initial vertical speed times time plus (1/2) a t^2
The two problems are connected by the time in the air, t
Yes, the two kinematic equations you have mentioned are correct. However, there are a couple of variations you can use depending on the information given.
In this problem, we are given the vertical distance (y-y₀) and the horizontal distance (x-x₀). We need to find the time the ball was in the air (t) and its initial vertical velocity (V₀y) or speed (since it's horizontally launched).
Since the ball rolls off the edge of the tabletop horizontally, its initial vertical velocity (V₀y) is zero. Therefore, the equation for the vertical motion becomes:
(y-y₀) = V₀yt - (1/2)gt²
Where:
- (y-y₀) is the vertical distance (1.80 m)
- V₀yt is the initial vertical velocity (which is zero since the ball is rolling horizontally)
- (1/2)gt² is the vertical displacement due to gravity (where g is the acceleration due to gravity, approximately 9.8 m/s²)
Since V₀yt is zero, the equation simplifies to:
1.80 m = - (1/2)gt²
Next, let's consider the horizontal motion. The horizontal distance traveled by the ball is given as 1.42 m, and the initial horizontal velocity (V₀x) is the velocity at the instant it left the table. We need to determine the time (t) for this equation. The horizontal motion equation is given as:
(x-x₀) = V₀xt
Where:
- (x-x₀) is the horizontal distance (1.42 m)
- V₀xt is the initial horizontal velocity (V₀x) multiplied by time (t)
We can now solve for time because we only have one unknown (t) in this equation:
1.42 m = V₀xt
Now, rearranging the equation for V₀x, we get:
V₀x = (x-x₀) / t
Substituting the given values into this equation, we have:
V₀x = 1.42 m / t
Since we don't know the value of t yet, we should solve the vertical motion equation for t and substitute it into V₀x equation. Let's solve for t:
1.80 m = - (1/2)gt²
Dividing by the negative factor, we get:
-3.60 m/g = t²
Taking the square root of both sides, we have:
t = √(-3.60 m/g)
But we know that time can't be negative in this context. Therefore, the equation tells us that the ball was in the air for a time of:
t = √(3.60 m/g)
Now, substitute this value of t into the horizontal motion equation for V₀x:
V₀x = 1.42 m / t
Calculate V₀x using the value of t you found and you will have the speed of the ball when it left the table.