range of function

xc630

Junior Member
Joined
Sep 1, 2005
Messages
164
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
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Sep 1, 2005
Messages
164
so the range would be 0>y</= 115 since 115 is the max height?
 

Gene

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Oct 8, 2003
Messages
1,904
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
Joined
Sep 1, 2005
Messages
164
The calculator shows the maximum height is 80 at 3 secs. So the new range is 0<x>80?
 

Gene

Senior Member
Joined
Oct 8, 2003
Messages
1,904
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.
 
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