A baseball is hit at ground level. The ball reaches its maximum height above ground level 2.6 s after being hit. Then 2.1 s after reaching its maximum height, the ball barely clears a fence that is 96.4 m from where it was hit. Assume the ground is level.
(a) What maximum height above ground level is reached by the ball?
m
(b) How high is the fence?
m
(c) How far beyond the fence does the ball strike the ground?
m
Horizontal equation:
96.4=VicosTheta*4.7
Vertical equation:
h=ViSinTheta*4.7-4.9(4.7^2)
one last equation: at max height, vy=0
0=ViSinTheta*2.6-4.9(2.6^2)
Lets see. Solve the third equation for VisinTheta, put that into the second equation, solve for h (answer b)
then max height:
hmax=visintheta(2.6)-4.9(2.6^2)
solve for hmax using the same expression you had in equation 3.
To find the answers to these questions, we can use the equations of motion for objects in free fall.
First, let's find the maximum height reached by the ball.
(a) The maximum height above ground level is reached when the vertical velocity of the ball becomes zero. We can use the equation of motion:
vf = vi + at
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.
At the maximum height, vf = 0, and vi is the initial vertical velocity of the ball. Since the ball is hit at ground level, vi = 0.
Therefore, the equation becomes:
0 = 0 + (-9.8) * t_max
Solving for t_max, we get:
t_max = 2.6 seconds
Now, to find the maximum height above ground level, we can use the equation:
h = vi * t + (1/2) * a * t^2
where h is the height, vi is the initial vertical velocity, a is the acceleration (which is -9.8 m/s^2 due to gravity), and t is the time.
Plugging in the values, we have:
h = 0 * 2.6 + (1/2) * (-9.8) * (2.6)^2
Simplifying, we get:
h = -12.78 meters
However, since we are interested in the maximum height above ground level, we take the absolute value:
h = 12.78 meters
So, the maximum height reached by the ball is 12.78 meters above ground level.
(b) To find the height of the fence, we need to know the time it takes for the ball to reach the fence. We can calculate this by subtracting the time it takes for the ball to reach its maximum height from the total time it takes to reach the fence:
t_f = total time - t_max
t_f = 2.1 seconds - 2.6 seconds
t_f = -0.5 seconds
(Note: The negative sign indicates that the ball reaches the fence 0.5 seconds before it reaches its maximum height.)
Now, we can use the equation of motion to find the height of the fence:
h = vi * t + (1/2) * a * t^2
Plugging in the values, we have:
h_fence = 0 * (-0.5) + (1/2) * (-9.8) * (-0.5)^2
Simplifying, we get:
h_fence = 1.225 meters
So, the height of the fence is 1.225 meters.
(c) To find how far beyond the fence the ball strikes the ground, we need to calculate the horizontal distance traveled by the ball in the time it takes to reach the fence.
Using the equation for horizontal distance:
d = vi * t
Since the initial horizontal velocity of the ball is constant, we can use the value at any time during the ball's flight. Let's use the time it takes to reach the fence:
d = vi * t_f = vi * (-0.5) = -0.5 * vi
To find the initial horizontal velocity (vi), we can use the horizontal distance traveled (96.4 meters) and the total time (2.1 seconds):
vi = d / t_total = 96.4 / 2.1 = 45.9048 m/s
Now, we can calculate the distance beyond the fence:
d_beyond_fence = -0.5 * 45.9048
Simplifying, we get:
d_beyond_fence = -22.9524 meters
However, since distance cannot be negative in this context, we take the absolute value:
d_beyond_fence = 22.9524 meters
Therefore, the ball strikes the ground 22.9524 meters beyond the fence.