Could u help with physics hw! i keep trying but i keep getting the wrong answer?
a 71g autographed baseball slides off of a 1.3 m high table and strikes the floor a horizontal distance of .83m away from the table.
1.how fast was it rolling on the table before it fell off?
ok well i already figured out the first question. which was 1.61
but for some reason i keep getting the
2nd question wrong.
can someone help!please
2.What was the ball's velocity just before it hit the floor?
That is, at what angle in the range -90° to +90° relative to the horizontal directed away from the table) did the ball hit the floor?
answer in units of °
It is still going 1.61 m/s horizontal if it was going that speed originally.
Vertical the speed is:
v = Vo + a t, the initial speed down, Vo, is zero
v = 0 - 9.8 t
what is t, the time to fall 1.3 m?
h = ho + Vo t + .5 a t^2
-1.3 = 0 + 0 - 4.9 t^2
t = .515 s
so
v = -9.8 * .515 = -5.05 m/s
speed = sqrt (-5.05^2 + 1.61^2)
= sqrt ( 25.5+2.59)
=5.3 m/s
tan angle to horizontal = 5.05/1.61 = 3.14
angle down from horizontal = 72.3 degrees down from horizontal
To find the ball's velocity just before it hit the floor, we can break down the problem into two components: the horizontal and vertical components of velocity.
First, let's find the time it takes for the ball to fall from the table to the floor. We can use the formula for the vertical displacement (Δy) of an object in free fall:
Δy = v_iy * t + 1/2 * a_y * t^2
In this case, the initial vertical position (y_i) is 1.3 m, the final vertical position (y_f) is 0 m, and the acceleration due to gravity (a_y) is -9.8 m/s^2 (negative because it acts downward). We need to find the time (t).
0 = v_iy * t + 1/2 * (-9.8 m/s^2) * t^2
Simplifying the equation:
0 = v_iy * t - 4.9 t^2
This is a quadratic equation. We can solve for t using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
In our case, the equation is in the form of at^2 + bt + c = 0, with:
a = -4.9
b = v_iy
c = 0
Using these values in the quadratic formula:
t = (-v_iy ± sqrt(v_iy^2 - 4*(-4.9)*0)) / (2*(-4.9))
Since the vertical velocity (v_iy) is positive (upward), the negative root doesn't make physical sense (negative time). Therefore, we take the positive root:
t = (-v_iy + sqrt(v_iy^2)) / (-9.8)
Next, we can find the horizontal displacement (Δx) of the ball during this time, using the formula:
Δx = v_ix * t
In this case, the initial horizontal position (x_i) is 0, and the final horizontal position (x_f) is 0.83 m. We need to find the horizontal velocity (v_ix).
0.83 m = v_ix * t
Now, we have two equations with two unknowns (v_iy and v_ix). We can solve these equations simultaneously to find their values.
1. From the first equation:
t = (-v_iy + sqrt(v_iy^2)) / (-9.8)
Replace t in the second equation:
0.83 m = v_ix * [(-v_iy + sqrt(v_iy^2)) / (-9.8)]
Solving for v_ix:
v_ix = (0.83 m * (-9.8)) / (-v_iy + sqrt(v_iy^2))
2. From the second equation:
0.83 m = v_ix * t
Replace t with the value found in equation 1:
0.83 m = v_ix * [(v_iy + sqrt(v_iy^2)) / (-9.8)]
Simplifying:
v_iy = (-9.8 * 0.83 m) / (v_ix + sqrt(v_ix^2))
Now, we have a system of equations:
v_iy = (-9.8 * 0.83 m) / (v_ix + sqrt(v_ix^2))
v_ix = (0.83 m * (-9.8)) / (-v_iy + sqrt(v_iy^2))
We can solve these equations simultaneously using a numerical method or by trial and error. By trying different values for v_ix, we can calculate v_iy. Then, we can calculate the angle (θ) using trigonometry:
θ = arctan(v_iy / v_ix)
Once we find the angle in radians, we can convert it to degrees:
Angle in degrees = (θ * 180°) / π
This will give us the angle at which the ball hits the floor relative to the horizontal, directed away from the table.