Respuesta :
Answer:
The distance traveled by the ball before it hits the ground is 255.10 meters.
Explanation:
Given that,
Initial velocity of the golf ball, u = 50 m/s
The angle of projection, [tex]\theta=45^{\circ}[/tex]
To find,
The distance traveled by the ball before it hits the ground.
Solution,
The distance covered by the projectile is called its horizontal range. It is given by the formula of range as :
[tex]R=\dfrac{v^2\ sin2\theta}{g}[/tex]
[tex]R=\dfrac{(50)^2\ sin2(45)}{9.8}[/tex]
R = 255.10 meters
So, the distance traveled by the ball before it hits the ground is 255.10 meters.
Answer:
255.1 m
Explanation:
initial velocity pf projection, u = 50 m/s
angle of projection, θ = 45°
The horizontal range is given by
[tex]R = \frac{u^{2}Sin2\theta }{g}[/tex]
[tex]R = \frac{50^{2}Sin90 }{9.8}[/tex]
R = 255.1 m
Thus, the ball travel 255.1 m horizontally before hitting the ground.
