Question

a ball is thrown and follows the parabolic path shown above. air friction is negligible. point q is the highest point on the path. points p and r are the same height above the ground.

1. how do the speeds of the ball at the three points compare?

(a) ( v_p < v_q < v_r ) (b) ( v_r < v_q < v_p ) (c) ( v_p = v_r < v_q ) (d) ( v_q < v_p = v_r )

1. which of the following diagrams best shows the direction of the acceleration of the ball at point p?

(a) ↙ (b) ↗ (c) ↘ (d) ↓ (e) acceleration is zero at point p

1. which of the following best indicates the direction of the velocity, if any, on the ball at point q?

(a) ↓ (b) → (c) ↘ (d) ↗ (e) there is no velocity at point q,

Explanation:

Question 8

Step1: Analyze vertical motion

The ball's vertical motion has acceleration \(g = 9.8\ m/s^{2}\) (downward). At point \(Q\), the vertical component of velocity \(v_{yQ}=0\). For points \(P\) and \(R\) (same height), using \(v_{y}^{2}=v_{0y}^{2}- 2gh\), since \(h\) is the same for \(P\) and \(R\), the vertical - component of velocity has the same magnitude at \(P\) and \(R\) (\(v_{yP}=v_{yR}\)).

Step2: Analyze horizontal motion

Horizontal motion has no acceleration (\(a_x = 0\), because \(F_{x}=0\) as \(F_{air - friction}=0\)). So \(v_{xP}=v_{xQ}=v_{xR}\) (constant horizontal velocity). The speed \(v=\sqrt{v_{x}^{2}+v_{y}^{2}}\). At \(Q\), \(v_Q = v_x\). At \(P\) and \(R\), \(v_P=\sqrt{v_{x}^{2}+v_{yP}^{2}}\) and \(v_R=\sqrt{v_{x}^{2}+v_{yR}^{2}}\), and \(v_{yP}=v_{yR}\)

The only force acting on the ball is the gravitational force \(F = mg\) (since air - friction is negligible). According to Newton's second law \(F = ma\), so \(a=\frac{F}{m}=g\) (downward) at all points of the trajectory (including point \(P\))

The velocity of a projectile at any point is tangent to its path. At the highest point \(Q\) of the projectile's path, the vertical component of velocity \(v_y = 0\) (as the ball stops moving upward and is about to move downward), and the horizontal component of velocity \(v_x\) is non - zero (because there is no horizontal acceleration, \(a_x=0\)). So the velocity is horizontal (to the right as per the direction of the projectile's motion)

Answer:

\(v_Q

Question 9