Suppose this firework is set to explode 3 seconds after it is launched. At what height will this firework be when it explodes?

My equation is h(t)=-16t^2+120t+20
How do I do this?

h = 20 + 120 * 3 - 16 (9)

= 20 + 360 - 144

in feet where g = 32 ft/s^2

this is because in feet and seconds

acceleration of gravity, g = 32 ft/s^2

then velocity v = Vi - g t
where Vi is initial velocity and g = 32 ft/s^2

and height h is
h = Hi + Vi t - (1/2)g t^2
in this problem
h = Hi + Vi t - 16 t^2
where Hi = 20 and Vi = 120

Thanks Damon!

You are welcome.

To determine the height at which the firework will explode, you need to find the value of 'h' when 't' is equal to 3 seconds. Here's how you can calculate it using your equation:

1. Substitute the value of 't' (3 seconds) into the equation h(t) = -16t^2 + 120t + 20:
h(3) = -16(3)^2 + 120(3) + 20

2. Simplify the equation by performing the necessary calculations:
h(3) = -16(9) + 360 + 20
h(3) = -144 + 360 + 20
h(3) = 236

Therefore, the firework will be at a height of 236 units when it explodes, according to the given equation.