Atechnician is launching an aerial firework from 200 -foot tower .the fireworks upward velocity at launch will be 176 feet per second. professional fireworks timed to explode as they reach their highest point .how long dose it take the firework to reach its maximum height ? what is the height ?

Hello there!

Here we need to use the second Newton's law, F = m*a

Now, the only force acting on the firework is the force of gravity, then we have that the acceleration of the firework is equal to the gravity aceleration.

a = -g = - 32.2(ft/s^2)

now if we want to obtain the velocity, we need to integrate the acceleration over time; we get:

v(t) = - 32.2(ft/s^2)*t + c

where c is a constant of integration, in this case, is the initial velocity, and we know that the initial velocity is equal to 176 ft/s, then the velocity is:

v(t) = - 32.2(ft/s^2)*t + 176ft/s

Now for the position, we need to integrate this again:

p(t) = - (1/2)*32.2(ft/s^2)*t^2 + 176ft/s*t + k

where again, k is a constant of integration, and now is equal to the initial position, and we know that the initial position of the firework is the height of the tower, this is 200ft, then the position is:

p(t) = -16.1(ft/s^2)*t^2 + 176ft/s*t + 200ft

Ok! now we have all the equations that we need.

the first question is: How long does it take the firework to reach its maximum height?  

If we want to find this, we need to find the time where the velocity changes of sign ( this means that the firework "stops" going upwards)

v(t) = - 32.2(ft/s^2)*t + 176ft/s = 0

32.2(ft/s^2)*t = 176ft/s

t = (176/32.2)s = 5.46 seconds

the second question is: What is the height ?

Now we need to evaluate the postion equation in the time we just find, this is:

p( t = 5.46 s) = -16.1(ft/s^2)*(5.46)^2 + 176ft/s*(5.46) + 200ft = 681 ft

i wish i can but busy atm

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