A golf ball is thrown at an angle of 37 degree with horizontal with initial speed of 50 m/s. what is the maximum horizontal distance???

L=vₒ²sin2α/g=50²•sin74º/9.8=245.2 m

this question is multiple choice there are choices and this answer is not among them :(

Upward motion (from the start point to the highest point)

v=v(0y)-gt ,
v=0
t= v(0y)/g=v(0) •sinα/g
the horizontal range is
L=2•v(x) •t=2•v(0) •cosα•v(0) •sinα/g= =v²(0) •sin2 α/g.
This is correct answer (both formula and calculation).
Show your set of answers.

To find the maximum horizontal distance, we need to break down the initial velocity into its horizontal and vertical components.

Given:
Angle of projection (θ) = 37 degrees
Initial speed (u) = 50 m/s

1. Finding horizontal component of velocity (ux):
The horizontal component of velocity remains constant throughout the projectile motion. It can be found using trigonometry:
ux = u * cos(θ)
= 50 * cos(37)
≈ 40.13 m/s

2. Finding vertical component of velocity (uy):
The vertical component of velocity changes due to the effect of gravity. It can be found using trigonometry:
uy = u * sin(θ)
= 50 * sin(37)
≈ 30.10 m/s

3. Finding time of flight (T):
The time taken for the projectile to reach the maximum height and return to the same height is called the time of flight. This can be calculated using the vertical component of velocity:
T = (2 * uy) / g
= (2 * 30.10) / 9.8
≈ 6.12 s

4. Finding maximum horizontal distance (R):
The maximum horizontal distance traveled by the projectile can be found by multiplying the horizontal component of velocity by the time of flight:
R = ux * T
≈ 40.13 * 6.12
≈ 245.57 m

Therefore, the maximum horizontal distance traveled by the golf ball is approximately 245.57 meters.