During a baseball game, a batter hits a high
pop-up.
If the ball remains in the air for 6.65 s, how
high does it rise? The acceleration of gravity
is 9.8 m/s
2
.
Answer in units of m
CAN ANYONE HELP PLEASE ??????!!!
can u help me pleaseeeeee @Damon
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
To find the height the ball rises, we can use the equation of motion under constant acceleration.
The equation is:
h = v₀t + (1/2)gt²
Where:
h = height the ball rises
v₀ = initial velocity of the ball (in this case, 0 m/s as the ball starts from rest)
t = time the ball remains in the air (6.65 s)
g = acceleration due to gravity (9.8 m/s²)
Plugging in the values, we have:
h = (0)(6.65) + (1/2)(9.8)(6.65²)
h = 0 + (1/2)(9.8)(44.2225)
h = 0 + 215.9585
h ≈ 216 m
Therefore, the ball rises approximately 216 meters.
falls from top for 6.65/2 = 3.325 seconds
h = (1/2) (9.8) (3.325)^2