Question

Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle θ below the horizontal. the baseline from which the ball is served is 11.9 m from the net, which is 0.91 m high. what is the angle θ such that the ball just crosses the net? will the ball land in the service box, which has an outermost service line is 6.40 m from the net?

Answer 1

serving speed = 170 km/h

now we will convert it into m/s

$v = 47.22 m/s$

now it will have two components along x and y directions

$v_x^2 + v_y^2 = 47.22^2$

also in x and y directions we know

$2.5 - 0.91 = v_y*t + \frac{1}{2}gt^2$

$11.9 = v_x *t$

$1.59 = v_y*t + 4.9*t^2$

$1.59 = 11.9 tan\theta + 4.9*(\frac{11.9}{47.22})^2(1 + tan^2\theta)$

$1.59 = 11.9 tan\theta + 0.311 (1 + tan^2\theta)$

by solving above equation we have

$tan\theta = 0.107$

$\theta = 6.11 degree$

so angle is 6.11 degree below the horizontal

now we have

$v_y = 47.22 sin6.11 = 5.02 m/s$

$v_x = 47.22 cos6.11 = 46.95 m/s$

now time taken by ball to reach the ground will be

$2.5 = 5.02 * t + \frac{1}{2}gt^2$

$t = 0.37 s$

now the displacement in x direction will be

$x = v_x * t$

$x = 46.95 * 0.37 = 17.37 m$

now the distance from net will be

$d = 17.37 - 11.9 = 5.47 m$

So it will land inside the line