MHB Word problem: initial height of projectile

AI Thread Summary
The height of the cliff from which a ball is thrown is determined using the equation h = -5t^2 + 5t + 210. By substituting t = 0 into the equation, the initial height is calculated as h = 210 meters. The discussion confirms that the cliff's height is indeed 210 meters. Understanding the use of the equation is crucial for solving similar projectile motion problems. The correct interpretation of the initial conditions leads to the accurate conclusion about the cliff's height.
mathdrama
Messages
20
Reaction score
0
A ball is thrown from a cliff. The path of the ball modeled by the equation
h = -5t^2+ 5t + 210,
where h is the height, in metres, of the ball above the ground, and t is the time, in seconds, after it is thrown. How high is the cliff?

Not really sure how to do this problem. I know that one of the roots is -6, but I don't know how to use this information.
 
Mathematics news on Phys.org
Re: word problem

mathdrama said:
A ball is thrown from a cliff. The path of the ball modeled by the equation
h = -5t^2+ 5t + 210,
where h is the height, in metres, of the ball above the ground, and t is the time, in seconds, after it is thrown. How high is the cliff?

Not really sure how to do this problem. I know that one of the roots is -6, but I don't know how to use this information.

Since $t$ is the time after the ball is thrown, you have to set $t=0$ at the equation to find the height of the ball before it's thrown, so when it is still on the cliff.
 
Re: word problem

So to make sure I have this right...

Let t = 0
h = -5(0^2) + 5(0) + 210
h = 0 + 0 + 210
h = 210
Therefore, the cliff is 210 meters high.
 
Re: word problem

mathdrama said:
So to make sure I have this right...

Let t = 0
h = -5(0^2) + 5(0) + 210
h = 0 + 0 + 210
h = 210
Therefore, the cliff is 210 meters high.

Yes, it is right! (Yes)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top