A batted baseball leaves the bat with a velocity of 41.4 m/s at an angle of 51.2°. If the ball just barely clears the 3.05 m (10 ft) center field wall,
a. How long was the ball in the air?
___s
b. What must the range be from home plate to the center field wall?
___m
c. How high did it rise above the point where the bat struck it?
___m
d. The final velocity is vf =
___m/s @ __°
Vo = 41.4m/s[51.2o].
Xo = 41.4*Cos51.2 = 25.9 m/s.
Yo = 41.4*sin51.2 = 32.3 m/s.
a. Y = Yo + g*Tr.
0 = 32.3 - 9.8Tr, Tr = 3.29 s. = Rise time.
h = Yo*Tr + 0.5g*Tr^2.
h = 32.3*3.29 - 4.9*3.29^2 = 53.2 m.
h = ho - 0.5g*Tf^2.
3.05 = 53.2 - 4.9Tf^2,
4.9Tf^2 = 53.2 - 3.05 = 50.18,
Tf = 3.2 s. = Fall time.
Tr+Tf = 3.29 + 3.20 = 6.49 s. = Time in air.
b. Range = Xo*(Tr+Tf) = 25.9*6.49 = 168 m.
c. h = 53.2 m.(Part a).
d. Y^2 = Yo^2 + 2g*h.
Y^2 = 32.3^2 + 19.6(53.2-3.05).
Y = 45 m/s.
V = Sqrt(Xo^2+Y^2) = Sqrt(25.9^2+45^2) = 51.9 m/s[60.1o].
)
Correction:
d. y^2 = Yo^2 + 2g*h.
Y^2 = 0 + 19.6*(53.2-3.05) = 982.94, Y = 31.35 m/s.
V = Xo+Yi = 25.9 + 31.35i = 40.7 m/s.[50.4].
To solve the problem, we can break down the initial velocity into its horizontal and vertical components.
The horizontal component of the initial velocity (vxi) can be calculated using the formula:
vxi = v * cos(θ)
where v is the magnitude of the initial velocity (41.4 m/s) and θ is the angle at which the ball was hit (51.2°).
Substituting the values, we get:
vxi = 41.4 * cos(51.2°)
Now, let's calculate the vertical component of the initial velocity (vyi) using the formula:
vyi = v * sin(θ)
Substituting the values, we get:
vyi = 41.4 * sin(51.2°)
a. To find how long the ball was in the air, we need to calculate the time it takes for the ball to reach the ground. We can use the vertical component of the initial velocity and the formula for time of flight:
t = 2 * vyi / g
where g is the acceleration due to gravity (9.8 m/s²).
Substituting the values, we get:
t = 2 * (41.4 * sin(51.2°)) / 9.8
b. To find the range from home plate to the center field wall, we can use the formula for range:
R = vxi * t
Substituting the values for vxi and t, we get:
R = (41.4 * cos(51.2°)) * t
c. To find how high the ball rose above the point where the bat struck it, we need to calculate the maximum height reached by the ball. We can use the formula:
hmax = (vyi^2) / (2 * g)
Substituting the value for vyi, we get:
hmax = ((41.4 * sin(51.2°))^2) / (2 * 9.8)
d. The final velocity (vf) can be calculated using the formula:
vf = √((vxi^2) + (vyi^2))
Now let's substitute the values for vxi and vyi into the formula to find vf.