A baseball flew 19.6 m straight up into the air after being hit by a bat. How long should it take for the ball to return from its maximum height?
A. 1.02 s
B. 1.91 s
C. 2.00 s
D. 3.99 s
3.99
Well, I hate to break it to you, but this baseball doesn't have a watch, so it won't be able to tell you exactly how long it takes to come back down. However, if we use some physics, we can calculate the time it takes. When a projectile is launched straight up, it will take the same amount of time to reach its maximum height as it does to come back down. So, we just need to find the time it takes for the ball to reach its maximum height. We can use the equation t = 2v/g, where v is the initial vertical velocity and g is the acceleration due to gravity. In this case, the initial velocity is given as 19.6 m/s (or should I say, "speedy baseball"), and the acceleration due to gravity is approximately 9.8 m/s². Plugging in the values, we get t = 2(19.6) / 9.8 = 3.99 s. So, the answer is D. 3.99 s. Just don't ask the baseball to confirm it, because it might throw you a curveball.
To solve this problem, we can use the kinematic equation for vertical motion:
v = u + at
Where:
v = final velocity (0 m/s at the maximum height)
u = initial velocity (unknown)
a = acceleration (acceleration due to gravity = -9.8 m/s^2)
t = time taken to reach maximum height (unknown)
At the maximum height, the final velocity is 0 m/s, so we can rearrange the equation to solve for u:
0 = u + (-9.8)t
Rearranging the equation:
-9.8t = u
Since we know the ball was hit straight up, we know its initial velocity in the vertical direction is positive:
u = 19.6 m/s
Substituting this value into the equation:
-9.8t = 19.6
Solving for t:
t = 19.6 / -9.8
t = -2
Since time cannot be negative in this case, we disregard the negative value. Therefore, it should take 2 seconds for the ball to return from its maximum height.
Therefore, the correct answer is:
C. 2.00 s
To answer this question, we can use the concept of projectile motion. When an object is thrown vertically into the air, it follows a path determined by the force of gravity. The maximum height of the ball occurs when its vertical velocity becomes zero.
First, let's calculate the time it takes for the ball to reach its maximum height. We can use the equation:
v = u + at
Where:
v = final velocity (zero at maximum height)
u = initial velocity (the velocity at which the ball was hit)
a = acceleration due to gravity (-9.8 m/s^2)
t = time
Since the ball reaches a height of 19.6 m, we can set v = 0 and solve for t:
0 = u - 9.8t
Rearranging the equation and solving for t:
t = u / 9.8
Next, to calculate the time it takes for the ball to return from its maximum height, we need to double the time it took to reach that height. This is because the time taken to reach the maximum height is equal to the time taken to descend back to the starting point.
So, the total time is:
Total time = 2 * t
Now, let's substitute the value for t into the equation to find the total time:
Total time = 2 * (u / 9.8)
Finally, we can compare the calculated value to the answer choices provided:
A. 1.02 s
B. 1.91 s
C. 2.00 s
D. 3.99 s
From the calculation, we can determine the correct answer.
S = distance = 1/2*g*t^2
Solve for t.