During a baseball game, a batter hits a
pop-up to a fielder 82 m away.
The acceleration of gravity is 9.8 m/s2 .
If the ball remains in the air for 6.9 s, how
high does it rise?
Answer in units of m.
goes up for 6.9/2 = 3.45 seconds
initial speed up = Vi
v = Vi - 9.8 t
at top, v = 0
0 = Vi - 9.8 (3.45)
Vi = 33.8 m/s
h = Vi t - 4.9 t^2
= 33.8(3.45) - 4.9(3.45^2)
= 116.6 - 58.3
=58.3 meters
note, the horizontal proble, the 82 m range, has nothing to do with the problem so far.
THANK YOU
Its been ten years my guy I don't think they are gonna respond
Wait why didn't we separate the x and y?
LMAOOOOO Swindle I canttt
Well, before I answer that, I have a joke for you. Why don't scientists trust atoms? Because they make up everything! Now, let's get to the calculation.
To find the height the ball rises, we can use the formula:
h = (1/2) * g * t^2,
where h is the height, g is the acceleration of gravity, and t is the time the ball remains in the air.
Plugging in the values, we have:
h = (1/2) * 9.8 m/s^2 * (6.9 s)^2.
Simplifying this equation, we get:
h = 1/2 * 9.8 m/s^2 * 47.61 s^2.
Solving this equation, we find that the ball rises approximately 232.8 meters high.
So, the ball reaches a height of 232.8 meters. Don't worry, it won't reach the moon! It's just a pop-up in a baseball game.
To find the height the ball rises, we can use the equation of motion for vertical motion:
Δy = Vi*t + (1/2)*a*t^2
Where:
Δy = change in height (the height the ball rises)
Vi = initial vertical velocity (ideally, 0 m/s when the ball leaves the ground)
a = acceleration due to gravity (-9.8 m/s^2, as it acts downward)
t = time the ball remains in the air (6.9 s)
In this case, the ball starts from the ground, so Vi = 0 m/s. Therefore, the equation simplifies to:
Δy = (1/2)*a*t^2
Substituting the given values:
Δy = (1/2)*(-9.8 m/s^2)*(6.9 s)^2
Calculating this expression, we can determine the height the ball rises.