During a baseball game, a batter hits a high pop-up.
If the ball remains in the air for 6.41 s, how
high above the point where it hits the bat
does it rise? Assume when it hits the ground
it hits at exactly the level of the bat.
acceleration of gravity is 9.8 m/s
During a baseball game, a batter hits a high pop-up.
If the ball remains in the air for 6.41 s, how
high above the point where it hits the bat
does it rise? Assume when it hits the ground
it hits at exactly the level of the bat.
acceleration of gravity is 9.8 m/s
can someones pls help im currently doing the timed homework
time up equals time down
d = 1/2 * g * (6.41 / 2)^2
your g number is two sig fig ... so your answer should be also
To determine the height the ball rises above the point where it hits the bat, we can use the equations of motion.
The key equation to use in this case is the equation for vertical motion, which is given by:
h = ut + (1/2)gt^2
Where:
h is the vertical height
u is the initial velocity in the upward direction
g is the acceleration due to gravity
t is the time
Given:
Time of flight, t = 6.41 s
Acceleration due to gravity, g = 9.8 m/s^2
We need to find the initial velocity, u, so let's use the fact that when the ball reaches its maximum height, its final velocity in the upward direction is 0.
At the maximum height, v = 0 and u is the initial velocity.
Using the equation of motion for vertical velocity:
v = u + gt
Substituting the values:
0 = u + (9.8)(6.41)
Rearranging the equation:
u = -9.8 (6.41)
u = -62.718 m/s
Note: The negative sign indicates that the initial velocity is in the downward direction.
Now we have the initial velocity, u, we can use it to find the maximum height, h.
h = ut + (1/2)gt^2
Substituting the values:
h = (-62.718)(6.41) + (1/2)(9.8)(6.41)^2
Calculating this expression will give us the height the ball rises above the point where it hits the bat.