At serve a tennis player aims to hit the ball horizontally. What minimum speed is required for the ball to clear the 0.90 m high net about 16.0 m from the server if the ball is hit from a height2.10m?

Answer 1

There is nothing wrong with the approach above my answer, but it is really useful (especially on tests and quizzes and the AP) to know how to do this class of problem generally, so I will take the time to "lay it out" for you. Once you own this stuff, these questions are a snap!

There are several ways to "cast" the projectile equations, but for the "does the ball clear the wall/net/whatever, the "solved" parametric equations (to get a parabola in x and y) are best...

To do this...

Use the equation: height=h=1/2 at^2 + vyo*t +h0 (with x=vxo * t) where vyo is the initial velocity in the y direction and vxo is the initial x direction velocity...which, of course, without air resistance is constant)

Substituting to eliminate the parameter t...

h = -1/2 a (x/vxo)^2 + vyo * (x/vxo) + ho

is now the general equation for finding the height of a projectile at ANY value of x (horizontal distance from launch).

Any number of questions can now be handled easily with this equation. Note h is positive UP, so I put the negative sign in front of the acceleration term to remind you that it acts DOWNWARD.

For your problem ho=2.5 m, vyo=0, let's use a=10.0 m/s^2 and just clearing the net, sets x=15.0 m (and h=.90 m).

Now, we can solve for the horizontal velocity that "get's the job done".

.90 = -1/2 (10.0) (15.0/vxo)^2 + 2.50 which gives...

1.60 * 2/10 = 225/vxo^2 = .32 so vxo=SQRT(225/.32) m/s

vxo=SQRT(731) = about 27 m/s (I do not have a calculator in my study).

To determine the flight time and landing position...

Recall that the flight time will be the same as if the ball were dropped from 2.50 meters...

h= 1/2 at^2 and so...5/10 = .50 = t^2 so t=.70 sec (about)

Now use vx=27 m/s with t=.70 to finf the range..

dist=rate * time = 27 * .70 = about 19 meters from the point of service.