A tennis ball is struck such that it leaves the racket horizontally with a speed of 25.0 m/s. The ball hits the court at a horizontal distance of 19.0 m from the racket. What is the height of the tennis ball when it leaves the racket?

 

 

I don't want the answer please. I just want to know what formula to use as well as a little tutoring on what the formula means and what the objects in the formula are. Thanks! Klaynos rocks my socks! LOL.

First thing you need to do is work out how long it'll take to get 18m at 25m/s using:

 

t = d/s

 

This is just simply working out the time it takes to travel horizontally, there is no acceleration in this direction (ignoring air resistance)

 

Then you can use:

 

s=ut + 0.5*at^2

 

u = 0

a=-9.81

t = whatever you calculate

 

This is working out the distance (s) vertically where u is the initial vertical velocity, t is the time it takes to drop which is the same to travel the distance, and a is the acceleration due to gravity...

And the reason you can do what Klaynos said is that the vertical motion and horizontal motion are independent. Gravity only affects the vertical motion, so a tennis ball dropped from the same height would hit at the same time as the one in the problem.

i got a negative answer

is that right?

Yes because I defined gravity as negative you'd get a negative answer, that would be for moving from the hand to the ground, so I would say the height, which would normally be measured from the ground to the hand would be positive. It just depends how you define the +ve and -ve directions...

i didn't get the right answer...i feel stupid

i am craving mexican food.

so i dont know why i got it wrong

i got negative 8.2

i didn't get the right answer...i feel stupid

i am craving mexican food.

so i dont know why i got it wrong

i got negative 8.2

So I never figured out the answer to this question. I only have one more try. ---Klaynos...in USA we obviously use different letters in our equations. I don't know what u means. V in US means velocity...maybe you can clear this up for me?

OnlyThorns,

 

In Klaynos's equation:

 

s is distance

u is velocity

a is acceleration

t is time

 

these are typical variables used for these factors in Newtonian physics - even in north america but you can rewrite it as

 

d = vt + 0.5a*t^2

 

so solving for the horizontal motion (as Klaynos said we neglect air resistance so we assume the acceleration is 0... this makes the second half the equation go away).

 

19 m = 25 m/s (t)

t = 0.76 seconds

 

Now look at the vertical motion:

Put 0.76 seconds into the original equation. Initial velocity in the VERTICAL direction is zero so now the first half of the equation goes away. Put in 9.81 m/s^2 for acceleration due to gravity and solve for d. This will be the height the ball falls during the 0.76 seconds it took to move forward the 19 metres... and thus the height at which the racket contacted the ball.