Object is thrown upward from a height of 15 ft at an initial vertical velocity of 30 ft per second. How long will it take to hit the ground?

2.25 s.

Step-by-step explanation:

We'll begin by calculating the time taken for the object to get to the maximum height from the point of projection. This can be obtained as follow:

Initial velocity (u) = 30 ft/s

Final velocity (v) = 0 ft/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s² = 3.28084 × 9.8 = 32.15 ft/s²

Time (t₁) to reach the maximum height from the point of projection =?

v = u – gt₁ (since the object is going against gravity)

0 = 30 – (32.15 × t₁

0 = 30 – 32.15t₁

Collect like terms

0 – 30 = – 32.15t₁

– 30 = – 32.15t₁

Divide both side by – 32.15

t₁ = –30 / –32.15

t₁ = 0.93 s

Next, we shall determine the maximum height reached by the object from the point of projection.

This can be obtained as follow:

Initial velocity (u) = 30 ft/s

Final velocity (v) = 0 ft/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s² = 3.28084 × 9.8 = 32.15 ft/s²

Maximum height (h) reached from the point of projection =?

v² = u² – 2gh (since the object is going against gravity)

0² = 30² – (2 × 32.15 × h)

0 = 900 – 64.3h

Collect like terms

0 – 900 = – 64.3h

– 900 = – 64.3h

Divide both side by – 64.3

h = –900 / –64.3

h = 14 ft

Thus, the maximum reached by the object from the point of projection is 14 ft.

Next, we shall determine the height to which the of object is located from the maximum height reached to the ground. This can be obtained as follow:

Height (h₀) from which the object was projected = 14 ft

Maximum Height (h) reached from the point of projection = 14 ft

Height (hₗ) to which the of object is located from the maximum to the ground =?

hₗ = h₀ + h

hₗ = 14 + 14

hₗ = 28 ft

Thus, the height to which the of object is located from the maximum reached to the ground is 28 ft.

Next, we shall determine the time taken for the object to get to the ground from the maximum height reached. This can be obtained as follow:

Height (hₗ) to which the of object is located from the maximum to the ground = 28 ft

Acceleration due to gravity (g) = 9.8 m/s² = 3.28084 × 9.8 = 32.15 ft/s²

Time (t₂) taken for the object to get to the ground from the maximum height reached =?

hₗ = ½gt₂²

28 = ½ × 32.15 × t₂²

28 = 16.075 × t₂²

Divide both side by 16.075

t₂² = 28 / 16.075

Take the square root of both side

t₂ = √(28 / 16.075)

t₂ = 1.32 s

Finally, we shall determine the time take for the object to get to the ground from the point of projection. This can be obtained as follow:

Time (t₁) to reach the maximum height from the point of projection = 0.93 s

Time (t₂) taken for the object to get to the ground from the maximum height reached = 1.32 s

Time (T) take for the object to get to the ground from the point of projection =?

T = t₁ + t₂

T = 0.93 + 1.32

T = 2.25 s.

Therefore, the time take for the object to get to the ground from the point of projection is 2.25 s.

tangent

step-by-step explanation:

x² + 3x + 7 = 0       [ax² + bx + c]

a = 1

b = 3

c = 7

quadratic formula:

    (+/- is this ±)

plug in the numbers into the variables

your answer is a