A softball player moving 3.89 m/s
1 answer:
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Answer:
A) t = 4.40 s , B) v = 23.86 m / s , c) v_y = - 43.12 m / s , D) v = 49.28 m/s
Explanation:
This is a projectile throwing exercise,
A) To know the time of the stone in the air, let's find the time it takes to reach the floor
y = y₀ + t - ½ g t²
as the stone is thrown horizontally v_{oy} = 0
y = y₀ - ½ g t²
0 = y₀ - ½ g t²
t = √ (2 y₀ / g)
t = √ (2 95 / 9.8)
t = 4.40 s
B) what is the horizontal velocity of the body
v = x / t
v = 105 / 4.40
v = 23.86 m / s
C) The vertical speed when it touches the ground
v_y = - g t
v_y = 0 - 9.8 4.40
v_y = - 43.12 m / s
the negative sign indicates that the speed is down
D) total velocity just hitting the ground
v = vₓ i ^ + v_y j ^
v = 23.86 i ^ - 43.12 j ^
Let's use Pythagoras' theorem to find the modulus
v = √ (vₓ² + v_y²)
v = √ (23.86² + 43.12²)
v = 49.28 m / s
we use trigonometry for the angle
tan θ = v_y / vₓ
θ = tan⁻¹ (-43.12 / 23.86)
θ = -61