Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 31.3 m/s and at an angle of 36.7 ∘ above the horizontal. You can ignore air resistance.

A) At what two times is the baseball at a height of 9.00 m above the point at which it left the bat?
t1,2 = _____ s
B) Calculate the horizontal component of the baseball's velocity at each of the two times you found in part (a).
vh1,2 = _____ m/s
C) Calculate the vertical component of the baseball's velocity at each of the two times you found in part (a).
v v1,2 = _____ m/s
D) What is the magnitude of the baseball's velocity when it returns to the level at which it left the bat?
v = ____ m/s
E) What is the direction of the baseball's velocity when it returns to the level at which it left the bat?
= ______ degrees below the horizontal

Answer 1

A major leaguer hits a baseball and the question asks for the time, velocity components, velocity magnitude, and direction at different points of the ball's trajectory when it reaches a certain height.

Step-by-step explanation:

A) Let's consider the vertical motion of the baseball. The equation to determine the time it takes for an object to reach a certain height is:

h = v0t + 0.5gt2

Where:

• h is the height = 9.00 m
• v0 is the initial vertical velocity = 31.3 m/s * sin(36.7º)
• g is the acceleration due to gravity = 9.8 m/s2

By rearranging the equation, we can solve for t:

t2 + (2v0/g)t - (2h/g) = 0

Using the quadratic formula, we can plug in the values and solve for t. The two possible values of t will correspond to the two times when the ball is at a height of 9.00 m above the point at which it left the bat.

B) The horizontal component of velocity remains constant throughout the motion. Therefore, the horizontal component of velocity at each of the two times found in part (a) will be the same as the initial horizontal velocity, which is 31.3 m/s * cos(36.7º).

C) The vertical component of velocity changes due to the acceleration of gravity. At each of the two times found in part (a), the vertical component of velocity can be found using the equation:

vv1,2 = v0 + gt

Where:

• v0 is the initial vertical velocity = 31.3 m/s * sin(36.7º)
• g is the acceleration due to gravity = 9.8 m/s2
• t is the time at each height, which were found in part (a)

D) When the baseball returns to the level at which it left the bat, its vertical velocity will be -v0, which means it will be moving downward with the same magnitude as the initial velocity but in the opposite direction. The horizontal component of velocity remains unchanged. Therefore, the magnitude of the baseball's velocity when it returns to the level at which it left the bat will be the square root of the sum of the squares of the horizontal and vertical components, which can be calculated using the formula:

v = √(vh2 + vv2)

Where:

• vh is the horizontal component of velocity, which was found in part (b)
• vv is the vertical component of velocity when the ball returns to the level of the bat, which is -v0

E) The direction of the baseball's velocity when it returns to the level at which it left the bat can be determined using trigonometry. The angle below the horizontal can be found as:

θ = tan-1(vv/vh)

Where:

• vv is the vertical component of velocity when the ball returns to the level of the bat, which is -v0
• vh is the horizontal component of velocity, which was found in part (b)

Answer 2

A) t₁ = 0.56 s t₂ =3.26 s

B) vh₁=vh₂ = 25.1 m/s

C) v₁ = 13.2 m/s v₂ = -13.2 m/s

D) v = 31. 3 m/s

E) 36.7º below horizontal.

Step-by-step explanation:

A) As the only acceleration of the baseball is due to gravity, as it is constant, we can apply the kinematic equations in order to get times.

First, we can get the horizontal and vertical components of the velocity, as the movements along these directions are independent each other.

v₀ₓ = v* cos 36.7º = 31.3 m/s * cos 36.7º = 25.1 m/s

v₀y = v* sin 36.7º = 31.3 m/s * sin 36.7º = 18.7 m/s

As in the horizontal direction, movement is at constant speed, the time, at any point of the trajectory, is defined by the vertical direction.

We can apply to this direction the kinematic equation that relates the displacement, the initial velocity and time, as follows:

Δy = v₀y*t -1/2*g*t²

We can replace Δy, v₀y and g for the values given, solving a quadratic equation for t, as follows:

4.9*t²-18.7t + 9 = 0

The two solutions for t, are just the two times at which the baseball is at a height of 9.00 m above the point at which it left the bat:

t = 1.91 sec +/- 1.35 sec.

⇒ t₁ = 0.56 sec t₂= 3.26 sec.

B) As we have already told, in the horizontal direction (as gravity is always downward) the movement is along a straight path, at a constant speed, equal to the x component of the initial velocity.

⇒ vₓ = v₀ₓ = 25.1 m/s

C) In order to get the value of the vertical components at the two times that we have just found, we can apply the definition of acceleration (g in this case), solving for vfy, as follows:

vf1 = v₀y - g*t₁ = 18.7 m/s - (9.8m/s²*0.56 sec) = 13.2 m/s

vf₂ = v₀y -g*t₂ = 18.7 m/s - (9.8 m/s²*3.26 sec) = -13.2 m/s

D) In order to get the magnitude of the baseball's velocity when it returns to the level at which it left the bat, we need to know the value of the vertical component at this time.

We could do in different ways, but the easiest way is using the following kinematic equation:

vfy² - v₀y² = 2*g*Δh

If we take the upward path, we know that at the highest point, the baseball will come momentarily to an stop, so at this point, vfy = 0

We can solve for Δh, as follows:

Δh = v₀y² / (2*g) = (18.7m/s)² / 2*9.8 m/s² = 17.8 m

Now, we can use the same equation, for the downward part, knowing that after reaching to the highest point, the baseball will start to fall, starting from rest:

vfy² = 2*g*(-Δh) ⇒ vfy = -√2*g*Δh = -√348.9 = -18. 7 m/s

The horizontal component is the same horizontal component of the initial velocity:

vx = 25.1 m/s

We can get the magnitude of the baseball's velocity when it returns to the level at which it left the bat, just applying Pythagorean Theorem, as follows:

v = √(vx)² +(vfy)² = 31.3 m/s

E) The direction below horizontal of the velocity vector, is given by the tangent of the angle with the horizontal, that can be obtained as follows:

tg Ф = vfy/ vx = -18.7 / 25. 1 =- 0.745

⇒ Ф = tg⁻¹ (-0.745) = -36.7º

The minus sign tell us that the velocity vector is at a 36.7º angle below the horizontal.