Respuesta :
Answer:
The correct option is;
b. 7.94 m
Explanation:
The given parameters of the jump of the long jumper are;
The angle above the horizontal with which the long jumper leaves the ground, θ = 20.0°
The speed with which the long jumper leaves the ground, u = 11.0 m/s
The furthest horizontal distance the long jumper jumps, given that the motion is equivalent to that of a particle, is given by the formula for the range, R, of a projectile motion as follows;
[tex]R = \dfrac{u^2 \times sin \left (2 \cdot \theta \right )}{g}[/tex]
Where;
g = The acceleration due to gravity ≈ 9.8 m/s²
u = The initial velocity of the long-jumper = 11.0 m/s
θ = The angle of the direction above the horizontal the long-jumper jumps = 20.0°
Plugging in the values, gives;
[tex]R = \dfrac{(11.0 \ m/s)^2 \times sin \left (2 \times 20.0 ^{\circ} \right )}{9.8} = \dfrac{121 \ m^2/s^2 \times sin \left (40.0 ^{\circ} \right )}{9.8 \ m/s^2} \approx 7.94 \ m[/tex]
How far the long-jumper goes = The range, R, of the projectile motion ≈ 7.94 m.
