During a baseball game, a batter hits a high
pop-up.
If the ball remains in the air for 6.36 s, how
high does it rise? The acceleration of gravity
is 9.8 m/s
2
.
Answer in units of m.
Well, let's get this straight: the baseball batter hits a high pop-up and I'm here to provide some comic relief. So, the ball remains in the air for 6.36 seconds, and we want to know how high it rises. Since gravity is playing a part in this game with an acceleration of 9.8 m/s², let's crunch the numbers.
Now, we know that the ball is going up against gravity, so we need to take that into account. Using the formula:
h = 0.5 * g * t²
Where h is the height, g is the acceleration due to gravity, and t is the time in seconds. Plugging in the numbers:
h = 0.5 * 9.8 m/s² * (6.36 s)²
Calculating this, we get:
h ≈ 0.5 * 9.8 m/s² * 40.4496 s²
And finally, simplifying it gives us:
h ≈ 198.5208 m
So, according to my calculations, the ball rises approximately 198.5208 meters high. Keep aiming for the stars, batter! But please, don't forget to catch the ball when it comes down. Safety first!
To find the height the ball rises, we can use the kinematic equation:
height = (1/2) * acceleration * time^2
Given:
acceleration = 9.8 m/s^2 (acceleration due to gravity)
time = 6.36 s
Plugging in the values, we can calculate the height:
height = (1/2) * (9.8 m/s^2) * (6.36 s)^2
= 0.5 * 9.8 * (6.36)^2
= 0.5 * 9.8 * 40.4496
= 198.2352 m
Therefore, the ball rises to a height of 198.2352 meters.
if yalll didnt understand, here also b/c why not
vf-vi=at, so find that negative number then divide by two and it will give u |vi|
make it positive
vf =0 at the top, so
0^2=vi^2 + 2a ∆x
then
-(vi^2)/(2a){should be 19.6}= ∆x
time up equals time down ... so the ball takes (6.36 s / 2) to travel up (or down)
distance = 1/2 g t^2 = 1/2 * 9.81 m/s^2 * (6.36 s / 2)^2