Question

a tennis player strikes a ball when it is 2 m above the ground, with a velocity 18 m/s at 16 degrees above the horizontal. the net is 4 m away horizontally and it is 0.9 m high. what is the magnitude of the velocity (in m/s) when the ball is vertically above the net? |v| = ? 24.62 m/s 15.83 m/s 17.51 m/s 4.16 m/s

Explanation:

Step1: Find the initial horizontal and vertical velocities

The initial velocity $v_0 = 18$ m/s. The initial horizontal velocity $v_{0x}=v_0\cos\theta$ and the initial vertical velocity $v_{0y}=v_0\sin\theta$, where $\theta = 16^{\circ}$.
$v_{0x}=18\cos(16^{\circ})\approx17.3$ m/s
$v_{0y}=18\sin(16^{\circ})\approx4.96$ m/s

Step2: Find the time it takes to reach the net horizontally

The horizontal motion is a uniform - motion with $x = v_{0x}t$. Given $x = 4$ m and $v_{0x}=18\cos(16^{\circ})$, we can solve for $t$.
$t=\frac{x}{v_{0x}}=\frac{4}{18\cos(16^{\circ})}\approx0.23$ s

Step3: Find the vertical velocity at time $t$

The vertical - motion is a uniformly - accelerated motion with $v_y = v_{0y}-gt$, where $g = 9.8$ m/s².
$v_y=v_{0y}-gt=4.96-9.8\times0.23=4.96 - 2.254 = 2.706$ m/s

Step4: Find the magnitude of the velocity at the net

The magnitude of the velocity $v$ is given by $v=\sqrt{v_{x}^{2}+v_{y}^{2}}$. Since $v_x = v_{0x}\approx17.3$ m/s and $v_y = 2.706$ m/s.
$v=\sqrt{(17.3)^{2}+(2.706)^{2}}=\sqrt{299.29 + 7.32}=\sqrt{306.61}\approx17.51$ m/s

Answer:

17.51 m/s