During a baseball game, a batter hits a pop up to a fielder 65m away. The acceleration of gravity is 9.8m/s^2. If the ball remains in the air for 5.7s, how high does it rise? Answer in units of m.
The ball spends half of the 5.7 s going up, and half coming down. So, in the last 2.85 s, the vertical distance it falls is
H = (g/2)(2.85)^2 = 39.8 meters.
That is also the vertical distance that it rises.
The distance of the fielder is not needed. You could use it to obtain the horizontal velocity (11.4 m/s) and launch angle. Vertical initial velocity = g*2.85 s = 27.93 m/s
Launch angle = arctan (27.93/11.4) = 67.8 degrees
To find the height the ball rises, we can use the equation of motion for vertical motion:
s = ut + (1/2)at^2
Where:
s = displacement (height in this case, which is what we are trying to find)
u = initial velocity (m/s)
t = time (s)
a = acceleration (m/s^2)
Since the ball is hit straight up, the initial velocity (u) would be zero because at its peak, the ball momentarily stops before falling down.
Given:
u = 0 (m/s)
t = 5.7 s
a = -9.8 m/s^2 (negative because gravity acts downwards)
Using the equation of motion, we can simplify it to:
s = (1/2)at^2
Substituting the given values, we have:
s = (1/2) * (-9.8 m/s^2) * (5.7s)^2
Simplifying further:
s = (0.5) * (-9.8 m/s^2) * 32.49 s^2
s = -15.96 m^2/s^2
The height the ball rises is -15.96 m^2/s^2. However, negative values indicate a downward direction, so we take the magnitude of the value to get the actual height.
Therefore, the ball rises approximately 15.96 meters.