1. ## Equation

Here's an equation that I don't understand;
1. The formula to find the height of an object 't' seconds after it is thrown upwards is given by h=ut-5t2(squared)
where u is the initial speed with which the object is thrown upwards.

a) Write the equation for an object thrown upwards with an initial velocity of u=80
so, this is h=80t-5t2

b) At what times is h=0? (Put h=0 in the equation and solve for t)

It's the 'b' section that I'm having trouble with....

Thanks

2. $5t^2-80t=0 <=> 5=\frac{80}{t}$ What is t then?

3. Originally Posted by Tren301
Here's an equation that I don't understand;
1. The formula to find the height of an object 't' seconds after it is thrown upwards is given by h=ut-5t2(squared)
where u is the initial speed with which the object is thrown upwards.

a) Write the equation for an object thrown upwards with an initial velocity of u=80
so, this is h=80t-5t2

b) At what times is h=0? (Put h=0 in the equation and solve for t)

It's the 'b' section that I'm having trouble with....

Thanks
Hi Tren301,

Set h = 0 and factor the quadratic to find the two times.

$-5t^2+80t=0$

$-5t(t-16)=0$

$t=0 \:\:and \:\: t=16$

So, the height is 0 when t = 0. That's at the beginning of the toss.

The height is 0 when t = 16. That's after it hits the ground at the end of the toss.

I feel like maybe there should've been some constant in the equation, since
it would be hard to toss an object from ground level.