A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.2 s after being hit. Then 2.7 s after reaching its maximum height, the ball barely clears a fence that is 81.0 m from where it was hit. Assume the ground is level. (a) What maximum height above ground level is reached by the ball? (b) How high is the fence? (c) How far beyond the fence does the ball strike the ground?
Draw the trajectory of the ball and indicate the following points: starting point O; the max height A; the terminal point of trajectory B (where the ball hits the ground); C is the position where the ball is after 2.7 sec after the max position (the height of the fence); and point D which is located symmetrically to point C but on OA.
The time of motion from the initial (O) point to the terminal (b) point is
t= 2vₒ•sinα/g.
t(OA)= 3.2s=t/2 =2•v₀•sinα/2•g=
=v(oy)/g =>
v(oy) = g•t(OA) =9.8•3.2 = 31.36 m/s.
t(DA) = t(AC) = 2.7 s. =>
t(OD) =t(OA)-t(DA) =3.2-2.7= 0.5 s.
The height of the point D= the height of the point C= h = the height of the fence:
(b) h = v(oy) •t(OD) –gt(OD)²/2 = =31.36•0.5 – 9.8•0.5²/2 = 14.46 m.
(a) H-h=g•t(AC)²/2 = 35.72 m.
Max height H= h+35.72 = 14.46+32.72 = 50.18 m.
(c)
81 = L/2 + x =
=v(ox) •31.36/g +v(ox) •2.7,
v(ox) •[(31.36/9.8) +2.7] = 81,
v(ox) = 81•9.8/(31.36 +2.7•9.8) = 13.73 m/s.
X = 13.73•0.5= 6.86 m.
To answer this question, we can use the principles of projectile motion. Let's break down the problem into parts:
(a) To find the maximum height reached by the ball, we need to consider the vertical motion. We know that the time taken to reach the maximum height is 3.2 seconds. Since the ball was hit at ground level, the initial vertical velocity is zero. We can use the equation for vertical motion:
h = u*t + (1/2)*a*t^2
Where:
h is the maximum height
u is the initial vertical velocity (zero in this case)
t is the time taken to reach the maximum height (3.2 seconds)
a is the acceleration due to gravity (-9.8 m/s^2)
Plugging in the values:
h = 0*(3.2) + (1/2)*(-9.8)*(3.2)^2
Simplifying the equation gives:
h = -4.9 * (3.2)^2
Calculate the value of h to find the maximum height reached by the ball, which turns out to be 50.176 meters.
(b) To find the height of the fence, we need to consider the vertical motion again. After reaching its maximum height, the ball takes an additional 2.7 seconds to clear the fence. We can use the same equation as before, this time using the total time of 6.2 seconds (3.2 s reaching maximum height + 2.7 s clearing the fence):
h = u*t + (1/2)*a*t^2
But this time, we need to find the height when t = 6.2 s.
h = 0*(6.2) + (1/2)*(-9.8)*(6.2)^2
By calculating the value of h, we find that the ball clears the fence with a height of 64.934 meters.
(c) To find how far beyond the fence the ball strikes the ground, we need to consider the horizontal motion. The horizontal velocity remains constant throughout the motion. The horizontal distance the ball travels can be calculated using the equation:
d = v*t
Where:
d is the horizontal distance traveled
v is the horizontal velocity (which we assume to be constant)
t is the time taken to reach the ground after clearing the fence (2.7 seconds in this case)
We can find the horizontal distance by multiplying the horizontal velocity by the time:
d = v * 2.7
However, we don't have the value of the horizontal velocity. Since the ball was hit at ground level and the ground is level, we can assume that the horizontal velocity is the same for both the upward and downward motion. Therefore, we can use the equation:
d = u*t
Where:
d is the horizontal distance traveled
u is the initial horizontal velocity (which is the same as the horizontal velocity)
Plugging in the values:
d = u * 2.7
To find the horizontal distance beyond the fence, you need to know the horizontal velocity (or initial horizontal velocity). Unfortunately, the question does not provide that information.