a ball thrown vertically upwards with a speed of 20 m/s strikes a target after peneterating a distance of 0.01 in target.calculate the reterdation produced by target.
V^2 = Vo^2 + 2a*d.
a = (V^2-Vo^2)/2d =
d = 2.54cm/in * 0.01in 0.0254 cm = 2.54*10^-4 m.
V^2 = Vo^2 + 2a*d.
a = (V^2-Vo^2)/2d =(0-400)/5.08*10^-4 = -7.87*10*5 m/s^2.
To calculate the retardation produced by the target, we need to determine the time taken for the ball to reach the target.
First, let's determine the time it takes for the ball to reach the highest point of its trajectory (when its velocity becomes zero). We can use the kinematic equation:
v = u + at
Where:
v = final velocity (0 m/s at the highest point)
u = initial velocity (20 m/s)
a = acceleration (retardation produced by the target)
t = time
Rearranging the equation to solve for time (t), we have:
t = (v - u) / a
Since v = 0 m/s, the equation becomes:
t = (0 - 20) / a
Next, let's determine the time it takes for the ball to fall from the highest point to the target. The time it takes to reach the highest point is equal to the time it takes to fall from the highest point to the target. Let's denote this time as 2t (twice the time).
Now, to calculate the distance fallen (0.01 m), we can use the equation for distance:
s = ut + (1/2)at²
For the upward motion, the equation becomes:
0.01 m = 0 + (1/2)a(2t)²
0.01 m = at²
Substituting the value of "t" found earlier, we have:
0.01 m = a[(0 - 20) / a]²
Simplifying the equation, we get:
0.01 m = 400 / a
Rearranging the equation, we have:
a = 400 / 0.01
a = 40000 m/s²
Therefore, the retardation produced by the target is 40000 m/s².