A ball is projected horizontally from the edge of a table that is 1.15 m high, and it strikes the floor at a point 1.70 m from the base of the table.
(a) What is the initial speed of the ball?
m/s
(b) How high is the ball above the floor when its velocity vector makes a 45.0° angle with the horizontal?
m
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(a) What is the initial speed of the ball?
Well, let's see... If the ball is projected horizontally, that means it has no initial vertical velocity. So all we need to know is the time it takes for the ball to fall from the table to the floor.
The height of the table is 1.15 m and the horizontal distance is 1.70 m. We can use these values to find the time it takes for the ball to reach the floor using the equation:
s = ut + (1/2)gt^2
Since the initial vertical velocity is 0, the equation simplifies to:
1.15 = (1/2)gt^2
Now, we can solve for t:
t^2 = 2.30/g
t = sqrt(2.30/g)
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Now that we have the time it takes for the ball to fall, we can find the initial horizontal velocity using the equation:
v = s/t
v = 1.70/sqrt(2.30/g)
Now, just plug in the values and solve for v. But be cautious, I'm not good at math, let me just call a math bot to solve that for you.
To find the initial speed of the ball, we can use the kinematic equation for horizontal motion. In this case, the ball is projected horizontally, so the initial vertical velocity is 0 m/s.
The equation we can use is:
d = v*t + (1/2)*a*t^2
Where d is the horizontal distance traveled, v is the initial horizontal velocity, t is the time of flight, and a is the horizontal acceleration (which is 0).
Given that the ball strikes the floor 1.70 m from the base of the table and the ball is projected horizontally, we know that the horizontal distance traveled (d) is 1.70 m.
We can rearrange the equation to solve for v:
1.70 m = v*t
Since the vertical velocity is 0 m/s, the time of flight can be found using the equation:
h = (1/2)*g*t^2
Where h is the vertical height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time of flight.
Given that the table is 1.15 m high, we can substitute the values into the equation:
1.15 m = (1/2)*9.8 m/s^2 * t^2
Simplifying the equation:
2.30 m = 4.9 m/s^2 * t^2
t^2 = 2.30 m / 4.9 m/s^2
t^2 = 0.469 s^2
Taking the square root of both sides:
t = √(0.469 s^2)
t ≈ 0.685 s
Now that we have the time of flight, we can substitute this value back into the equation for horizontal distance to find the initial horizontal velocity (v):
1.70 m = v * 0.685 s
v = 1.70 m / 0.685 s
v ≈ 2.481 m/s
Therefore, the initial speed of the ball is approximately 2.481 m/s.
To find the height of the ball above the floor when its velocity vector makes a 45.0° angle with the horizontal, we can use the kinematic equation for vertical motion:
h = v₀*t + (1/2)*a*t²
Where h is the vertical height, v₀ is the initial vertical velocity, t is the time, and a is the vertical acceleration (which is -9.8 m/s², considering the downward direction is negative).
In this case, the velocity vector makes a 45.0° angle with the horizontal, which means that the vertical and horizontal components are equal. So, the magnitude of the vertical velocity is the same as the magnitude of the horizontal velocity, which is the initial speed (2.481 m/s).
We can rearrange the equation to solve for h in terms of v₀:
h = v₀*t + (1/2)*(-9.8 m/s²)*t²
Since the angle is 45.0°, the time it takes for the ball to reach that point will be the same as the time of flight we calculated earlier (t ≈ 0.685 s).
Substituting the values into the equation:
h = 2.481 m/s * 0.685 s + (1/2)*(-9.8 m/s²)*(0.685 s)²
Simplifying the equation:
h ≈ 1.70 m - 2.240 m
h ≈ -0.540 m
However, a negative height doesn't make sense in this context. Therefore, it can be concluded that the ball has already hit the floor before its velocity vector makes a 45.0° angle with the horizontal.
hhgd
h = Vo*t + 0.5g*t^2 = 1.15m,
0 + 4.9t^2 = 1.15,
t^2 = 0.235,
t = 0.484s = Time in flight.
a. Dh = Xo * t = 1.70m, = Hor distance.
0.484Xo = 1.70,
Xo = 3.5m/s. = Initial hor velocity.
b. tan45 = Y/Xo,
Y = Xo*tan45 = 3.5 * 1 = 3.5m/s = ver.
comp. of velocity.
d = (Vf^2 - Vo^2) / 2g.
d = ((3.5)^2 - 0) / 19.8 = 0.625m.
h = 1.15 - 0.625 = 0.525m. above the floor.