The impulse delivered to the ball by the floor after rebounding with the same speed and angle is; Δp = 2.74 N.s
We are given;
Mass of basketball; m = 0.6 kg
Speed; v = 5.4 m/s
Angle to the vertical; θ = 65°
We want to find the impulse if the ball rebounds with the same speed and angle above.
Now, as the x-component of the momentum remains constant, the impulse would be equal to the change in the y-component of the balls' momentum.
Thus;
Δp = m[(v_f * cos θ) - (v_i * cos θ)]
Earlier we saw that we were given the rebound speed to be the same with the initial speed but however, as the y-axis is pointing upwards, it means that the initial velocity will be negative.
Thus;
v_f = 5.4 m/s
v_i = -5.4 m/s
Plugging in the relevant values into the impulse equation gives;
Δp = 0.6[(5.4 * cos 65) - (-5.4 * cos 65)]
Δp = (0.6 × cos 65) × (5.4 + 5.4)
Δp = 2.74 N.s
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The impulse delivered to the ball by the floor after the rebound is 5.87 Ns.
The given parameters;
The impulse delivered to the ball is the change in the linear momentum of the ball in vertical direction.
The magnitude of the impulse delivered to the ball by the floor is calculated as;
[tex]J = \Delta P = m(v_f - v_0)\\\\J = m(vsin\theta - vsin\theta)\\\\the \ final \ speed \ occured \ in \ opposite \ direction;\\\\J = m(vsin\theta + vsin\theta)\\\\J = m(2vsin\theta)\\\\J = 2mv \times sin(\theta)\\\\J = 2\times 0.6 \times 5.4 \times sin(65)\\\\J = 5.87 \ Ns[/tex]
Thus, the impulse delivered to the ball by the floor after the rebound is 5.87 Ns.
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