Amajor leaguer hits a baseball so that it leaves the bat at a speed of 32.7 m/s and at an angle of 34.6° above the horizontal. you can ignore air resistance. a)at what two times is the baseball at a height of 11.5 m above the point at which it left the bat?
b)calculate the horizontal component of the baseball's velocity at an earlier time calculated in part a.
c)calculate the vertical component of the baseball's velocity at an earlier time calculated in part a.
d)calculate the horizontal component of the baseball's velocity at a later time calculated in part a.
e)calculate the vertical component of the baseball's velocity at a later time calculated in part a.
f)what is the magnitude of the baseball's velocity when it returns to the level at which it left the bat?
g)what is the direction of the baseball's velocity when it returns to the level at which it left the bat?

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:

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.

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

conservation of mass

mass before something happens = mass afterwards