The height of a projectile is launched upward at a speed of 32 feet/per second from a height of 48 feet. how long will it take the projectile to hit the ground

The general equation of one-dimensional motion is

h(t) = h(0) + v(0)*t + (1/2)*a*t^2. Adapted to one-dimensional vertical motion,

h(t) = h(0) + v(0)*t + (1/2)*(-32.2 ft/sec^2), where -32.2 ft/sec^2 is the acceleration due to gravity, in the English system of measurement.

Further adapting this equation to solving the current problem:

h(t) = 48 ft + (32.2 ft/sec)*t - (16.1 ft/sec^2)*t^2

We want to know how long it will take for the projectile to reach its maximum height and then hit the ground. To answer this, set h(t) = 0 and solve for t:

0 = 48 ft + (32.2 ft/sec)*t - (16.1 ft/sec^2)*t^2

This is a quadratic equation, with a = -16.1, b = 32.2 and c = 48.

-(-32.2) plus or minus √( (-32.2)^2 - 4(-16.1)(48) )

t =

2(-16.1)

This will give us two roots; one negative and one positive. This positive root represents the length of time required for the particle to reach its maximum height and then hit the ground. Omit the negative root.

hello :

70y^8 + 30y^6 = 2 ( 35 y^8 + 15y^6)

the lenght is : 35 y^8

the width is : 15y^6

b

step-by-step explanation:

it have 5 sides and 5 identicle angle.the rest does not have the same property as b.

brainliest