A golf ball is hit with an initial velocity of 50 m/s at an angle of 45° above the horizontal. How far will the ball travel horizontally before it hits the ground?

To find the horizontal distance traveled by the golf ball before it hits the ground, we can use the following steps:

Step 1: Resolve the initial velocity into its horizontal (x) and vertical (y) components.
The initial velocity, v = 50 m/s.
The angle of projection, θ = 45°.

The horizontal component, vx = v * cos(θ)
= 50 * cos(45°)
= 50 * (√2 / 2)
= 25√2 m/s

The vertical component, vy = v * sin(θ)
= 50 * sin(45°)
= 50 * (√2 / 2)
= 25√2 m/s

Step 2: Calculate the time of flight of the ball.
The time it takes for the ball to hit the ground can be found using the vertical component of velocity and the acceleration due to gravity.

Using the equation:
y = vy * t + (1/2) * g * t^2

Where:
y = Initial vertical displacement = 0 (since the ball starts from the ground)
vy = Vertical component of velocity = 25√2 m/s
g = Acceleration due to gravity = 9.8 m/s^2
t = Time of flight

By substituting the values in the equation, we get:
0 = (25√2) * t + (1/2) * (9.8) * t^2

Rearranging the equation becomes a quadratic equation:
(1/2) * (9.8) * t^2 + (25√2) * t = 0

This equation can be easily solved to find the values of 't'.

Step 3: Find the horizontal distance traveled.

The horizontal distance is the product of horizontal velocity and the time of flight:
Horizontal distance, dx = vx * t

Substituting the values, we get:
dx = (25√2) * t

This will give us the horizontal distance traveled by the golf ball before it hits the ground.

To find how far the golf ball will travel horizontally before it hits the ground, you can use the kinematic equations of motion.

First, let's break down the initial velocity into its horizontal and vertical components. The initial velocity of 50 m/s makes an angle of 45° with the horizontal.

The horizontal component of the velocity (Vx) can be found using the equation:
Vx = V * cos(θ)
where V is the initial velocity (50 m/s) and θ is the angle (45°).

Vx = 50 * cos(45°)
Vx = 50 * 0.707 (rounded to three decimal places)
Vx ≈ 35.355 m/s

Now that we have the horizontal component of the velocity, we can use it to calculate the time it takes for the ball to hit the ground.

For an object in free fall, the time it takes to hit the ground can be found using the equation:
t = 2 * Vy / g
where Vy is the vertical component of the velocity and g is the acceleration due to gravity (approximately 9.8 m/s²).

Since the ball is thrown upwards at an angle, the initial vertical component of the velocity (Vy) can be found using the equation:
Vy = V * sin(θ)
where V is the initial velocity and θ is the angle.

Vy = 50 * sin(45°)
Vy ≈ 35.355 m/s

Using this value, we can calculate the time it takes for the ball to hit the ground:
t = 2 * 35.355 / 9.8
t ≈ 7.23 seconds (rounded to two decimal places)

Now that we know the time it takes for the ball to hit the ground, we can calculate the horizontal distance traveled (d).

The horizontal distance traveled can be calculated using the equation:
d = Vx * t
where Vx is the horizontal component of the velocity and t is the time.

d = 35.355 * 7.23
d ≈ 255.647 meters (rounded to three decimal places)

Therefore, the golf ball will travel approximately 255.647 meters horizontally before hitting the ground.

hyt

Vo = (50m/s,45deg.).

Xo = hor. = 50cos45 = 35.36m/s.

Yo = ver. = 50sin45 = 35.36m/s.

t(up) = (Vf - Yo) / g,
t(up) = (0 - 35.36) / -9.8 = 3.61s.

t(dn) = t(up) = 3.61s.

T = t(up) + t(dn)=3.61 + 3.61=7.22s. =
Time in flight.

Dh = Xo * T = 35.36 * 7.22 = 255m. =
Hor. distance.

Alternate Method:

Dh = Vo^2 * sin2A / g.
A = 45 deg.
Vo = 50m/s.