A fireworks shell is shot straight up with the initial velocity of 152 feet per second.

Question

A fireworks shell is shot straight up with the initial velocity of 152 feet per second. It's height s feet and t seconds is given by the equation s(t)=152t-16t^2. If the shell is designed when it reaches its maximum height, how long after being fired, and at what height, will the fireworks appear in the sky? Give the first answer in two decimal places and the second answer as the whole number.

Answer 1

s(t) = 152t - 16t^2

 

This is an equation describing constant acceleration motion.  It is of the form

 

s(t) = s0 + v0*t +(1/2)*a*t^2,

 

where s0 = starting height = 0 feet, v0 = initial vertical velocity = 152 feet/s, and a = acceleration of gravity = -32 feet/s^2

 

Maximum height is reached when the velocity drops to zero:

 

v(t) = v0 + a*t = 152 - 32*t

 

152 - 32*t = 0

 

t = 152/32 s = 4.75 s

 

h(4.75) = 152*4.75 - 16*(4.75)^2 = 361 feet

 

h_max = 361 feet, reached in 4.75 s