# range of function

#### xc630

##### Junior Member
This is a function prb. im having trouble finding the range

From a platform 35m above the gorund a ball is thrown upward with an initial speed of 30m/s. The approximate height of the ball above the ground t seocnds later is given by h (t)= 35 +30t-5t^2.

a) after how many secs does the ball hit the ground.
using -b/2a i got 6 secs.

b) what is the domain- using a grpahing calc i got -1<x>7

c) what is the range? I do not know how to do this. I would appreciate any help.

G

#### Guest

##### Guest
Many ways to solve the range depending on what level of maths you have available...
if you have learnt about dh/dt, then at max height the dh/dt = 0. Solve for this and you will find the t value for max. height. (min height =0)

or taking a science view if the total flight time is 6 seconds, then the max height occurs at half of this (3 seconds). Use this value to find max. height.

Hope this gets it going.

#### xc630

##### Junior Member
so the range would be 0>y</= 115 since 115 is the max height?

#### Gene

##### Senior Member
Hmmmm, one grows confused.
a) good formula, wrong application. a is negative so the parabola opens downward. -b/2a is the vertex formula and is the t coordenent of the <u>maximum</u> height.

b) Negative time is not part of the problem. The range starts when the ball is thrown and goes till it hits the ground. The 7 is O.K. but it is [0, 7] The calc showed where it was one second before it was thrown :?: The flight time is 7 seconds.

c) Plug the answer from a) into the original equation for the maximum height. The calc should have shown what the answer is. Not 115.

#### xc630

##### Junior Member
The calculator shows the maximum height is 80 at 3 secs. So the new range is 0<x>80?

#### Gene

##### Senior Member
Right numbers. I don't care for your notation though. Either
0<u><</u>h<u><</u>80
or
h=[0,80]
are the two ways I am familiar with.