A ball is thrown up at the edge of a 448 foot cliff. The ball is thrown up with an initial velocity of 72 feet per second.
Its height measured in feet is given in terms of time t, measured in seconds by the equation
h=-16t^2+72t+448

a. How high will the ball go and how long does it takes to reach that height? __feet __seconds
b. How long does it take the ball to come back to the ground ? ___seconds

Question

A ball is thrown up at the edge of a 448 foot cliff. The ball is thrown up with an initial velocity of 72 feet per second. 
Its height measured in feet is given in terms of time t, measured in seconds by the equation 
h=-16t^2+72t+448

a. How high will the ball go and how long does it takes to reach that height?   __feet  __seconds 
b. How long does it take the ball to come back to the ground ? ___seconds

campbell_st · Answer

there are several ways to do this.. 

1. calculus... 

2. algebra.. 

for me, the simplier method is algebra.. 

find the line of symmetry of the curve, and the max height is on the line of symmetry

in general for a parabola 

$$ax^2 + bx + c$$

the line of symmetry is 

$$x = \frac{-b}{2a}$$

so in your question you are finding t, and a = -16 and b = 72

substitute to find the time (t) when the object gets to the max height. 

subsitute t into the original equation to find the max height

campbell_st · Answer

for (b)

you need to let h(t) = 0

that is then does the ball have no height.... 


so you need to solve 

$$0 = -16t^2 + 72t + 448$$

or 

$$0 = -8(2t^2 - 9t - 56)$$

only consider the positive answer as time can't be negative

anonymous · Answer

thank you! =)