Question

Problem 7.12 Part A 0.145-kg baseball pitched horizontally at 37.1 /s strikes a bat and is popped straight up to a If the contact time is 2.90 ms, calculate the magnitude of the average force on the ball during the contact. height of 59.5 m. Express your answer using two significant figures F- Submit My Answers Give Up Part B Calculate the direction of the average force on the ball during the contact. Express your answer using two significant figures. θ= above horizontal Submit My Answers Give Up Provide Feedback Continue

Answer 1

given , mass of baseball, m = 0.145 kg

contact time , t = 2.90 ms = 2.90 * 10^-3 s

initial velocity of baseball is 37.1 m/s horizontally i.e.

$\vec{u}=37.1\hat{i} \ m/s$

after striking bat it moves vertically upto the height of 59.5 m i.e. h = 59.5 m

under the influence of acceleration due to gravity acting downward , g = - 9.8 m/s²

at height h = 59.5 m velocity of ball will be zero

Applying third equation of motion

$0^{2}=v^{2} + 2 gh$

$0^{2}=v^{2} + 2 \times \left ( -9.8 \right )\times 59.5$

$v=\sqrt{1166.2} \ m/s =33.15\ m/s \ upward$

$\therefore \vec{v}= 33.15\hat{j} \ m/s$

$\therefore \vec{F} = \frac{m\left ( \vec{v}-\vec{u} \right )}{t}=\frac{0.145\left ( 33.15\hat{j}-37.1\hat{i} \right )}{2.9\times 10^{-3}}$

$\vec{F} = 50\times \left ( 33.15\hat{j}-37.1\hat{i} \right )$

$\vec{F} = -1855\hat{i} +1657.5\hat{j}$

$F=\sqrt{\left ( -1855 \right )^{2}+\left ( 1657.5 \right )^{2}} = 2487.64 \ N$

$\theta = \tanh \frac{F_{y}}{F_{x}}=\tanh \frac{1657.5}{-1855} = 46.42^{0}\ above \ horizontal\ clockwise$

$OR \ \theta = 180^{0} -46.42^{0}=133.58^{0} \ above \ horizontal\ counterclockwise$