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
To find the maximum horizontal distance, we need to split the initial velocity into its horizontal and vertical components.
The horizontal component (Vx) of the velocity is given by: Vx = V * cos(theta)
where V is the initial speed (50 m/s) and theta is the angle of projection (37 degrees).
Substituting the values, we get: Vx = 50 * cos(37)
The vertical component (Vy) of the velocity is given by: Vy = V * sin(theta)
where V is the initial speed (50 m/s) and theta is the angle of projection (37 degrees).
Substituting the values, we get: Vy = 50 * sin(37)
Now, we need to find the time (t) it takes for the golf ball to reach its maximum height. At maximum height, the vertical component of the velocity becomes zero.
Using the kinematic equation: Vy = Vy0 - g * t
where Vy is the vertical component of the velocity, Vy0 is the initial vertical component of the velocity, g is the acceleration due to gravity (-9.8 m/s^2), and t is the time.
Substituting the values, we get: 0 = 50 * sin(37) - 9.8 * t
Simplifying the equation, we get: t = (50 * sin(37)) / 9.8
Now, we can find the maximum horizontal distance (D) traveled by the golf ball using the horizontal component of velocity (Vx) and the time taken (t).
Using the equation: D = Vx * t
Substituting the values, we get: D = (50 * cos(37)) * [(50 * sin(37)) / 9.8]
Simplifying the equation, we get: D = (50^2 * cos(37) * sin(37)) / 9.8
Evaluating the equation, we find that the maximum horizontal distance is approximately 191.8 meters.