Question

A volleyball is spiked over the net at an angle of 50° below the horizontal with a net acceleration of 3.6 m/s2. What is the horizontal force of the ball if it weighs 24 N?

Answer 1

First, we can find the mass of the volleyball from the definition of the weight:

$$W=mg,$$ $$m=\dfrac{W}{g}=\dfrac{24\ N}{9.8\ \dfrac{m}{s^2}}=2.4\ kg.$$

Then, we can find the net force acting on the ball from the Newton's Second Law of Motion:

$$F_{net}=ma_{net}=2.4\ kg\cdot 3.6\ \dfrac{m}{s^2}=8.64\ N.$$

Finally, we can find the horizontal force acting on the ball (projection of the net force onto x-axis):

$$F_{h}=F_{net}cos\theta=8.64\ N\cdot cos50^{\circ}=5.55\ N.$$

Answer:

$F_{h}=5.55\ N.$