A ball was thrown with a velocity of 150 ft/ s at angle of 45 degrees above the horizontal.

a)Find the range and the time of flight.
b)What is the maximum height reached?

a) Let the angle be A and the range R.

g = 32.2 ft^2/s.
T = time of flight

R = (V^2/(2g)) sin(2A)
T = 2*V*sinA/g
Note that 2A = 90 degrees so
sin(2A) = 1. Thus R = V^2/(2g) @ A=45
b) Max. height H = (V sinA)^2/(2g)

Thanks drwls...

a) If the ball was thrown at an angle of 45 degrees above the horizontal with a velocity of 150 ft/s, we can determine the range and time of flight.

To find the range, we use the formula:

Range = (Velocity^2 * sin(2 * angle)) / g,

where g is the acceleration due to gravity (32.2 ft/s^2).

Range = (150^2 * sin(2 * 45)) / 32.2,
Range = (22500 * sin(90)) / 32.2,
Range = (22500 * 1) / 32.2,
Range ≈ 697.51 ft.

To find the time of flight, we use the formula:

Time of Flight = (2 * velocity * sin(angle)) / g.

Time of Flight = (2 * 150 * sin(45)) / 32.2,
Time of Flight = (300 * √2) / 32.2,
Time of Flight ≈ 4.37 s.

b) To find the maximum height reached, we use the formula:

Maximum Height = (velocity^2 * sin(angle)^2) / (2 * g).

Maximum Height = (150^2 * sin(45)^2) / (2 * 32.2),
Maximum Height = (22500 * 0.5) / (64.4),
Maximum Height ≈ 175.43 ft.

So, the maximum height reached by the ball is approximately 175.43 ft.

To find the range and the time of flight, we can use the following formulas:

Range (R) = (velocity^2 * sin(2θ)) / g

Time of flight (T) = (2 * velocity * sin(θ)) / g

where:
- velocity is the initial velocity of the ball (150 ft/s),
- θ is the angle of projection (45 degrees),
- g is the acceleration due to gravity (32.2 ft/s^2).

a) Range (R):
Using the formula for range, we can substitute the given values:

R = (150^2 * sin(2 * 45)) / 32.2

R = (22500 * sin(90)) / 32.2

Since sin(90) = 1, the formula reduces to:

R = (22500 * 1) / 32.2

R ≈ 699.38 ft

Therefore, the range is approximately 699.38 ft.

Time of Flight (T):
Using the formula for time of flight, we can substitute the given values:

T = (2 * 150 * sin(45)) / 32.2

T = (300 * (sqrt(2) / 2)) / 32.2

T = (150 * sqrt(2)) / 32.2

T ≈ 4.33 s

Therefore, the time of flight is approximately 4.33 seconds.

b) To find the maximum height reached, we can use the formula:

Maximum height (H) = (velocity^2 * sin^2(θ)) / (2 * g)

Using the given values, we have:

H = (150^2 * sin^2(45)) / (2 * 32.2)

H = (22500 * (sqrt(2) / 2)^2) / (2 * 32.2)

H = (22500 * 2/4) / (2 * 32.2)

H = 22500 / (4 * 32.2)

H ≈ 88.22 ft

Therefore, the maximum height reached is approximately 88.22 ft.

To find the range and the time of flight of the ball, we need to analyze the motion of the projectile.

a) Range:
The range is the horizontal distance traveled by the projectile before it hits the ground. We can calculate it using the formula:

Range = (Initial Velocity * Time of Flight * cos(Angle))

Given:
- Initial Velocity (v₀) = 150 ft/s
- Angle (θ) = 45 degrees

Firstly, let's convert the angle from degrees to radians:
Angle (θ) = 45 degrees * (π/180) = π/4 radians

Now, we can calculate the time of flight (t):

Time of Flight (t) = (2 * Initial Velocity * sin(Angle)) / Acceleration due to gravity

Acceleration due to gravity (g) is approximately 32.17 ft/s².

Time of Flight (t) = (2 * 150 ft/s * sin(π/4)) / 32.17 ft/s²

Using a calculator, we can calculate the value of t, which is approximately 4.32 seconds.

Now, substituting the values into the range formula:

Range = (150 ft/s * 4.32 s * cos(π/4))

Again, using a calculator, we can calculate the value of the range, which is approximately 461.94 feet.

Therefore, the range of the ball is approximately 461.94 feet, and the time of flight is approximately 4.32 seconds.

b) Maximum height reached:
To find the maximum height reached by the ball, we can use the formula:

Maximum Height = (Initial Velocity * sin(Angle))² / (2 * Acceleration due to gravity)

Substituting the values:

Maximum Height = (150 ft/s * sin(π/4))² / (2 * 32.17 ft/s²)

Again, using a calculator, we can calculate the value of the maximum height, which is approximately 114.53 feet.

Therefore, the maximum height reached by the ball is approximately 114.53 feet.