# Trig Help Word Problem

• Jun 19th 2008, 08:12 AM
tricey36
Trig Help Word Problem
I just don't get these type of problems at all. Please help if at all possible the problem is as follows:

An astronaut on the moon tosses a ball upwards; when will the ball initially be 35 feet above the lunar surface if the ball's height(feet) follows:
-2.6t^2 + 32.1t + 5.5 and t is in seconds from release?
• Jun 19th 2008, 08:20 AM
TheEmptySet
Quote:

Originally Posted by tricey36
I just don't get these type of problems at all. Please help if at all possible the problem is as follows:

An astronaut on the moon tosses a ball upwards; when will the ball initially be 35 feet above the lunar surface if the ball's height(feet) follows:
-2.6t^2 + 32.1t + 5.5 and t is in seconds from release?

The height $h(t)=-2.6t^2+32.1t+5.5$

Since we are given a height we set $h(t)=35$ to get

$35=-2.6t^2+32.1t+5.5 \iff 0=-2.6t^2+32.1t-29.5$

We can multiply the equation by 10 to eliminate the decimals

$0=-26t^2+321t-295$ we can then factor this to get (you could also use the quadratic formula)

$-(t-1)(26t-295)=0$ so t=1 or $t=\frac{295}{26}\approx 11.3$

Good luck.
• Jun 19th 2008, 08:23 AM
Chris L T521
Quote:

Originally Posted by tricey36
I just don't get these type of problems at all. Please help if at all possible the problem is as follows:

An astronaut on the moon tosses a ball upwards; when will the ball initially be 35 feet above the lunar surface if the ball's height(feet) follows:
-2.6t^2 + 32.1t + 5.5 and t is in seconds from release?

First, I don't see how we need trig here...

All you need to do here is set $35=-2.6t^2+32.1t+5.5$ and solve for t [this is a quadratic that can be easily solved using the quadratic equation].

Note that you'll get two values both of which should be positive. The smaller value of t tells you the time it takes to initially reach the height of 35 feet (before reaching the maximum height). The second t value tells you when it reaches a height of 35 after it has reached its maximum height and starts to fall back to the surface.

If you still have questions, feel free to ask.

--Chris