Question

a ball is dropped from the top of a very tall building and falls with negligible air resistance. the instantaneous power delivered to the ball by the force of gravity when the ball has a velocity of v is p. what is the power delivered to the ball when its velocity is 2v? a 4p b 2p c p d 1/2 p

Explanation:

Step1: Recall the formula for instantaneous power

The formula for instantaneous power \(P\) is \(P = \vec{F}\cdot\vec{v}\). When a ball is in free - fall, the force acting on it due to gravity is \(F = mg\) (where \(m\) is the mass of the ball and \(g\) is the acceleration due to gravity), and the velocity \(\vec{v}\) is in the direction of the force of gravity. So, \(P=mgv\) (since the angle \(\theta\) between \(\vec{F}\) and \(\vec{v}\) is \(0^{\circ}\) and \(\cos\theta = 1\)).

Step2: Calculate the power when the velocity is \(2v\)

Let \(P_1\) be the power when the velocity is \(v\), so \(P_1=mgv\). When the velocity is \(v_2 = 2v\), the power \(P_2\) is given by \(P_2=mgv_2\). Substitute \(v_2 = 2v\) into the power formula: \(P_2=mg(2v)\). Since \(P = mgv\), then \(P_2 = 2(mgv)=2P\).

Answer:

B. \(2P\)