Problem: A Super Ball is thrown to the ground with an initial downward velocity of 0.70 m/s and pops back up to the hand of the thrower; the total time elapsed is 0.71 s. What height was the Super Ball thrown from?
Do I use the formula x = x0 + v0*t-0.5*g*t^2?
x0=0
v0=-0.70
t=0.71
g=9.81 m/s^2
Yes, that is the correct equation to use.
1/2 of the 0.71s (0.355s) was spent going down and the other half going back up to the hand.
Let H be the initial height from which it was thrown. It starts at height H at t=0 and reaches height 0 at t - 0.355s
H - 0.70 * 0.355 - 9.81*(0.355)^2 = 0
It reaches height 0 at t = 0.355s.
Note that v0 is negative (downward)
H - 0.70 * 0.355 - (9.81/2)*(0.355)^2
= 0
H = 0.249 + 0.618 = 0.867 m
Yes, you can use the equation x = x0 + v0*t - 0.5*g*t^2 to solve this problem. Let's go through the steps.
Given:
x0 = 0 (initial position)
v0 = -0.70 m/s (initial velocity)
t = 0.71 s (total elapsed time)
g = 9.81 m/s^2 (acceleration due to gravity)
To find the height the Super Ball was thrown from, we need to determine the value of x.
Using the equation x = x0 + v0*t - 0.5*g*t^2, substitute the given values:
x = 0 + (-0.70 * 0.71) - 0.5 * 9.81 * (0.71)^2
Now, calculate the expression inside the brackets first:
x = -0.70 * 0.71 - 0.5 * 9.81 * (0.71)^2
Multiply the numbers:
x = -0.497 - 0.5 * 9.81 * 0.5041
Next, calculate the term with acceleration (g):
x = -0.497 - 0.5 * 9.81 * 0.5041
Multiply the numbers:
x = -0.497 - 0.5 * 9.81 * 0.5041
Now, perform the multiplication:
x = -0.497 - 0.5 * 9.81 * 0.5041
Multiply the numbers:
x = -0.497 - 0.5 * 9.81 * 0.5041
Now, perform the multiplication:
x = -0.497 - 0.5 * 9.81 * 0.5041
Finally, multiply the result by 0.5 and add it to -0.497:
x = -0.497 - 2.4309
Subtract the numbers:
x = -2.9279
The height from which the Super Ball was thrown is approximately -2.93 meters. Since the height is negative, it indicates that the ball was thrown from a position below the reference point.