Respuesta :
Answer:
18.7 m
Explanation:
[tex]v_{o}[/tex] = initial speed of the shot = 13 m/s
θ = angle of launch from the horizontal = 47 deg
Consider the motion along the vertical direction
[tex]v_{oy}[/tex] = initial velocity along vertical direction = 13 Sin47 = 9.5 m/s
[tex]a_{y}[/tex] = acceleration along the vertical direction = - 9.8 m/s²
y = vertical displacement = - 1.80 m
t = time of travel
Using the kinematics equation
[tex]y=v_{oy} t+(0.5)a_{y} t^{2}[/tex]
- 1.80 = (9.5) t + (0.5) (- 9.8) t²
t = 2.11 s
Consider the motion along the horizontal direction
x = horizontal displacement of the shot
[tex]v_{ox}[/tex] = initial velocity along horizontal direction = 13 Cos47 = 8.87 m/s
[tex]a_{x}[/tex] = acceleration along the horizontal direction = 0 m/s²
t = time of travel = 2.11 s
Using the kinematics equation
[tex]x=v_{ox} t+(0.5)a_{x} t^{2}[/tex]
x = (8.87) (2.11) + (0.5) (0) (2.11)²
x = 18.7 m
The horizontal distance traveled by the shotput when launched at a velocity of 13 m/s is 18.7 m.
What is projectile motion?
A projectile motion is a curved motion that is projected near the surface of the earth such that it falls under the influence of gravity only.
Given to us
The initial velocity of the shotput, u = 13 m/s
The angle from the horizontal, θ = 47.0°
The height from which shotput was thrown h = 1.80 m
As we know that when the shotput will be thrown it will be in a projectile motion, therefore, the distance that will be covered by the shot put will be the range of the projectile motion,
[tex]R = \dfrac{u\ cos\theta}{g} [u\cdot sin\theta + \sqrt{u^2sin^2\theta+2gh}][/tex]
Substitute the values,
[tex]R = \dfrac{13\ cos47^o}{9.81} [13\cdot sin47 + \sqrt{13^2(sin47)^2+2(9.81)(1.80)}][/tex]
R = 18.7 m
Hence, the horizontal distance traveled by the shotput when launched at a velocity of 13 m/s is 18.7 m.
Learn more about Projectile Motion:
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