Answer:
a) At time t = 0.780 s and t = 3.01 s, the ball is at 11.5 m above the point at which it left the bat.
b) The horizontal component of the ball´s velocity at any given time is 26.9 m/s.
c) The vertical component of the velocity at t = 0.5 s is 13.7 m/s.
d) The horizontal component of the velocity is constant, 26.9 m/s.
e) The vertical component of the velocity at t = 3.5 s is -15.7 m/s
f) The magnitude of the baseball´s velocity when it returns to the level at which it left the bat is 32.7 m/s
g) The direction of the velocity vector is 34.7° below the horizontal when the ball reaches the level at which it left the bat.
Explanation:
The position and velocity of the ball is given by the following vectors:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
v = (v0 · cos α, v0 · sin α + g · t)
Where:
r = position vector at time t.
x0 = initial horizontal position.
t = time.
α = launching angle.
y0 = initial vertical position
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
v = velocity vector at time t.
Please, see the attached figure for a description of the problem. Notice that the origin of the frame of reference is located at the launching point so that x0 and y0 = 0.
a) When the ball is at a height of 11.5 m, the y-component of the position vector is 11.5 m. Using the equation for the vertical position, we can obtain the time:
y = y0 + v0 · t · sin α + 1/2 · g · t²
11.5 m = 32.7 m/s · t · sin 34.6° - 1/2 · 9.8 m/s² · t²
0 = -4.9 m/s² + 32.7 m/s · t · sin 34.6° - 11.5 m
t = 0.780 s and t = 3.01 s
At time t = 0.780 s and t = 3.01 s, the ball is at 11.5 m above the point at which it left the bat.
b) The horizontal component of the velocity is constant and does not depend on time:
vx = v0 · cos α
vx = 32.7 m/s · cos 34.6° = 26.9 m/s
c) The vertical component of the ball´s velocity is given by this equation:
vy = v0 · sin α + g · t
Let´s calculate the vertical velocity at t = 0.5 s
vy = 32.7 m/s · sin 34.6° - 9.8 m/s² · 0.5 s
vy = 13.7 m/s
d) The horizontal velocity is constant (ignoring air resistance). Then:
vx = v0 · cos α
vx = 32.7 m/s · cos 34.6° = 26.9 m/s
e) Let´s calculate the vertical component of the velocity vector at t = 3.5 s.
vy = 32.7 m/s · sin 34.6° - 9.8 m/s² · 3.5 s
vy = -15.7 m/s (the velocity is negative because the ball is falling at this time).
f) According to our frame of reference, the level at which the ball leaves the bat is at y = 0 m. Then, using the equation for the vertical position, we can calculate the time at which the ball returns to y = 0 m:
y = y0 + v0 · t · sin α + 1/2 · g · t²
0 = 32.7 m/s · t · sin 34.6° - 1/2 · 9.8 m/s² · t²
0 = t (32.7 m/s · sin 34.6° - 1/2 · 9.8 m/s² · t)
0 = 32.7 m/s · sin 34.6° - 1/2 · 9.8 m/s² · t
- 32.7 m/s · sin 34.6°/ -4.9 m/s² = t
t = 3.79 s
Now, we can calculate the y-component of the velocity at that time:
vy = v0 · sin α + g · t
vy = 32.7 m/s · sin 34.6° - 9.8 m/s² · 3.79 s
vy = -18.6 m/s
The x-component of the velocity was already calculated:
vx = 26.9 m/s
The magnitude of velocity vector will be:
[tex]|v| = \sqrt{(26.9 m /s)^{2} + (-18.6 m/s)^{2}}= 32.7 m/s[/tex]
Using trigonometry we can calculate the angle below the horizontal of the velocity vector ( see figure):
cos θ = vfx / vf
cos θ = 26.9 m/s / 32.7 m/s
θ = 34.7°
Then, the direction of the velocity vector is 34.7° below the horizontal when the ball reaches the level at which it left the bat.
