4. A baseball is hit straight up into the air at an initial velocity of 35 m/s. How long does it take for the baseball to reach its highest position?
v = 35 - 9.81t
it reaches its top height when the ball stops rising. (v=0)
To find out how long it takes for the baseball to reach its highest position, we can use the kinematic equation for displacement:
s = ut + (1/2)at²
Where:
s = displacement (0, since the baseball will reach its highest position)
u = initial velocity (35 m/s)
t = time taken
a = acceleration (in this case, acceleration due to gravity, which is approximately -9.8 m/s²)
Since the baseball is hit straight up, the acceleration due to gravity acts in the opposite direction, so we take it as negative.
Plugging in the known values, the equation becomes:
0 = (35 m/s) * t + (1/2) * (-9.8 m/s²) * t²
We can simplify this equation to:
0 = 35t - 4.9t²
Rearranging the equation to solve for time, we get:
4.9t² - 35t = 0
Now we can factor out t:
t(4.9t - 35) = 0
We have two possible solutions:
1) t = 0, which is not meaningful in this context since it represents the initial time.
2) 4.9t - 35 = 0
4.9t = 35
t = 35 / 4.9
Evaluating this expression, we find:
t ≈ 7.14 seconds
Therefore, it takes approximately 7.14 seconds for the baseball to reach its highest position.