Respuesta :
Answer:
The speed at which she threw the marshmallow is approximately 4.76 m/s
Explanation:
The given in formation are;
The range in which the marshmallow was thrown = 2 m (The location of her friend)
The angle at which the marshmallow was thrown = 30 degrees
The horizontal range is given by the following formula;
[tex]R = \dfrac{u^2 \times sin(2\cdot \theta)}{g}[/tex]
Where;
R = The range = 3 m
u = The initial velocity
θ = The angle at which the marshmallow was thrown = 30°
g = The acceleration due to gravity = 9.81 m/s²
By substitution, we have;
[tex]2 = \dfrac{u^2 \times sin(2\times 30)}{9.81} = \dfrac{u^2 \times sin(60)}{9.81}[/tex]
[tex]\therefore u^2 = \dfrac{ 2 \times 9.81}{sin(60)} = \dfrac{u^2 \times sin(60)}{9.81} \approx 22.66[/tex]
u ≈ √(22.66) ≈ 4.76 m/s
u ≈ 4.76 m/s
The speed at which she threw the marshmallow ≈ 4.76 m/s.
