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Schach [20]
3 years ago
9

An object is thrown upward from the top of a 144​-foot building with an initial velocity of 128 feet per second. The height h of

the object after t seconds is given by the quadratic equation h equals negative 16 t squared plus 128 t plus 144. When will the object hit the​ ground?
Physics
1 answer:
ankoles [38]3 years ago
8 0

Answer:

After 9 seconds the object reaches ground.

Step-by-step explanation:

We equation of motion given as h = -16t²+128t+144,

We need to find in how many seconds will the object hit the ground,

That is we need to find time when h = 0

                        0 = -16t²+128t+144

                       16t²-128t-144= 0

                       t=\frac{-(-128)\pm \sqrt{(-128)^2-4\times 16\times (-144)}}{2\times 16}\\\\t=\frac{128\pm \sqrt{25600}}{32}\\\\t=\frac{128\pm 160}{32}\\\\t=9s\texttt{ or }t=-1s

Negative time is not possible, hence after 9 seconds the object reaches ground.

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Answer:

60 m/s

Explanation:

a =  \frac{vf - vi}{t}

5.0 =  \frac{vf - 0}{12}

5.0 \times 12 = vf

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6 0
3 years ago
n outer space, a constant net force with a magnitude of 140 N is exerted on a 32.5 kg probe initially at rest. A) What accelerat
musickatia [10]

Answer:

a) 4.31 m/s²

b) 215.5 m

Explanation:

a) According to Newton's first law of motion

The net force applied to particular mass produced acceleration, a, according to

F = ma

F = 140 N

m = 32.5 kg

a = ?

140 = 32.5 × a

a = 140/32.5 = 4.31 m/s²

b) Using the equations of motion, we can obtain the distance travelled by the object in t = 10 s

u = initial velocity of the probe = 0 m/s (since it was initially at rest)

a = 4.31 m/s²

t = 10 s

s = distance travelled = ?

s = ut + at²/2

s = 0 + (4.31×10²)/2 = 215.5 m

7 0
3 years ago
An object is dropped from rest from a 70.6 m tower. Air resistance is negligible. After 0.32 seconds, what is magnitude and dire
dem82 [27]

Answer:

<em>1,378.9ms²</em>

Explanation:

Given the following

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Time t = 0.32secs

Initial velocity = 0m/s

Required

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Using the equation of motion

S = ut+1/2at²

Substitute

70.6 = 0+1/2a(0.32)²

70.6 = 0.0512a

a = 70.6/0.0512

a = 1,378.9

<em>Hence the acceleration is 1,378.9ms²</em>

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Similarities between land and sea breezes?
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Answer:

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Explanation:

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