A projectile with an initial velocity of 48 feet per second is launched from a building 190 feet tall. The path of the projectile is modeled using the equation h(t) = –16t2 + 48t + 190.

t = 5.25 seconds

Step-by-step explanation:

Given that,

A projectile with an initial velocity of 48 feet per second is launched from a building 190 feet tall. The path of the projectile is modeled using the equation:

It is assumed to find the time when the projectile will hit the ground. When the projectile hit the ground, its height is equal to 0 such that,

It forms a quadratic equation such that,

So, the projectile will hit the ground at t = 5.25 seconds.

50-38=12

step-by-step explanation:

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation: