Find the horizontal range of a baseball that leaves the bat at an angle of 63 degrees, with the horizontal with an initial velocity of 160 km/hr. discard air resistance

do the vertical problem

Vi = 160,000/3600 sin 63 = 44.4 m/s sin 63
= 39.6 m/s
how long to reach top where v = 0 ?
v = Vi - 9.8 t
0 = 39.6 - 9.8 t
t = 4.04 seconds to top
that means in the air for 8.08 seconds

now the horizontal problem
goes for 8.08 seconds at 44.4 cos 63 m/s
you can do that

163 Meters

Why did the baseball go to the circus?

Because it wanted to be a high-flying acrobat instead of sticking to sports!

To find the horizontal range of a projectile, we need to consider the horizontal component of its initial velocity and the time it takes to reach the ground.

Given:
Initial angle (θ) = 63 degrees
Initial velocity (v) = 160 km/hr

Step 1: Convert the velocity to m/s
To convert km/hr to m/s, divide by 3.6 (since 1 km/hr = 1000 m / 3600 s).

v = (160 km/hr) / (3.6) = 44.44 m/s (rounded to 2 decimal places)

Step 2: Calculate the horizontal component of the velocity.
The horizontal component (Vx) of the velocity is given by:
Vx = v * cos(θ)

Vx = 44.44 m/s * cos(63 degrees)

Vx ≈ 44.44 m/s * 0.447 = 19.86 m/s (rounded to 2 decimal places)

Step 3: Find the time of flight.
To find the time of flight, we need to determine the total time it takes for the projectile to reach the ground. Since there is no vertical acceleration (in the absence of air resistance), the time of flight will be the same for both the upward and downward parts of the motion.

The formula to calculate the time of flight (t) is given by:
t = (2 * v * sin(θ)) / g

where g is the acceleration due to gravity, approximately 9.8 m/s^2.

t = (2 * 44.44 m/s * sin(63 degrees)) / 9.8 m/s^2

t ≈ 7.24 seconds (rounded to 2 decimal places)

Step 4: Calculate the horizontal range (R).
The horizontal range (R) can be defined as the horizontal distance traveled by the projectile during the time of flight (t).

R = Vx * t

R ≈ 19.86 m/s * 7.24 seconds

R ≈ 143.77 meters (rounded to 2 decimal places)

Therefore, the horizontal range of the baseball, in the absence of air resistance, is approximately 143.77 meters.

To find the horizontal range of the baseball, we need to break down its initial velocity into horizontal and vertical components. The horizontal component of velocity remains constant throughout the motion because there is no external force acting on it.

Given:
Initial velocity (v) = 160 km/hr
Launch angle (θ) = 63 degrees

First, we need to convert the initial velocity from km/hr to m/s since the unit of time in physics calculations is usually in seconds.

1 km/hr = 1000 m / (60 × 60) s = 0.2778 m/s

Therefore, the initial velocity (v) in m/s is 160 × 0.2778 = 44.45 m/s.

Now, we can calculate the horizontal component of velocity (v_x) using the trigonometric function, cosine.

v_x = v × cos(θ)

Substituting the given values, we have:

v_x = 44.45 × cos(63)

Using a calculator, we find that v_x is approximately 19.45 m/s.

Next, we can calculate the horizontal range (R) using the formula:

R = (v_x^2 × sin(2θ)) / g

where g is the acceleration due to gravity (approximately 9.8 m/s²).

Substituting the values into the formula, we have:

R = (19.45^2 × sin(2 × 63)) / 9.8

Calculating this expression, we find that the horizontal range is approximately 133.15 meters.

So, the horizontal range of the baseball, neglecting air resistance, is approximately 133.15 meters.