A student wants to launch a tennis ball horizontally just over a 6.0 m high fence and have it land at approximately 20.0 m beyond the fence. What speed would the tennis ball need to have to land at this approximate distance?

Explanation:

Time taken for vertical displacement will be equal to time taken for horizontal displacement .

Let it be t

for vertical fall , initial velocity u = 0

h = ut + 1/2 gt²

6 = 0 + .5 x 9.8 t²

t = 1.1065 s

horizontal displacement = horizontal velocity x time

20 = horizontal velocity x 1.1065

horizontal velocity = 18 m /s approx.

seeing the graph would

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.