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Andrei [34K]
3 years ago
10

In a football game, the running back takes a handoff and begins running toward midfield at 3.21 yards/s . As he moves through hi

s linemen, he accelerates at a rate of 1.71 yards/s^2 to a velocity of 7.54 yards/s before he is tackled. How far did he run after he got the ball?
Physics
1 answer:
Inessa [10]3 years ago
8 0

Given Information:

Initial speed = u = 3.21 yards/s

Acceleration = α  = 1.71 yards/s²

Final speed = v = 7.54 yards/s

Required Information:

Distance = s = ?

Answer:

Distance = s = 13.61

Explanation:

We are given the speeds and acceleration of the runner and we want to find out how much distance he covered before being tackled.

We know from the equations of motion,

v² = u² + 2αs

Where u is the initial speed of the runner, v is the final speed of the runner, α is the acceleration of the runner and s is the distance traveled by the runner.

Re-arranging the above equation for distance yields,

2αs = v² - u²

s = (v² - u²)/2α

s = (7.54² - 3.21²)/2×1.71

s = 46.55/3.42

s = 13.61 yards

Therefore, the runner traveled a distance of 13.61 yards before being tackled.

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A stone was dropped off a cliff and hit the ground with a speed of 88 ft/s. What is the height (in feet) of the cliff
igor_vitrenko [27]

Answer:

the height (in feet) of the cliff is 121 ft

Explanation:

A stone hit the cliff with

speed, v = 88 ft/s

Acceleration, a= 32 ft/s^2

initial speed, u = 0 ft/s

height is h.

To solve this problem we will apply the linear motion kinematic equations, Equation of motion describes change in velocity, depending on the acceleration and the distance traveled

so, writing the formula of Equation of motion:

v^2 - u^2 = 2*a*h

substituting the appropriate values,

(88)^2 - 0 = 2*32* h

h=(88)^2 / 64

h= 121 ft

hence

the height (in feet) of the cliff is 121 ft

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3 0
1 year ago
A basketball player can leap upward 0.43 m. how long does he remain in the air? use an acceleration due to gravity of 9.80 m/s2
MArishka [77]
From the equations of linear motion,
v² = u² + 2as where v is the final velocity, u is the initial velocity and a is the gravitational acceleration, and s is the displacement,
Thus, v² = u² -2gs, but v=0
hence, u² = 2gs
                = 2×9.81×0.43
                = 8.4366
            u = √8.4366
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Hence the initial velocity is 2.905 m/s
 Then using the equation v= u +gt .
Therefore, v = u -gt. (-g because the player is jumping against the gravity)
but, v = 0
Thus, u= gt
Hence, t = u/g
              = 2.905/9.81
              = 0.296 seconds


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3 years ago
3. During a tug-of-war, Team A pulls with a
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Answer:

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Explanation:

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2 years ago
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A bullet is fired horizontally at a height of 1.3 meters and a velocity of 950 m/s. How long was the bullet in the air?
seropon [69]

Answer:

<em>The bullet was 0.52 seconds in the air.</em>

Explanation:

<u>Horizontal Motion </u>

It occurs when an object is thrown horizontally with a speed v from a height h.

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To calculate the time the object takes to hit the ground, we use the following equation:

\displaystyle t=\sqrt{\frac{2y}{g}}

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The bullet is fired horizontally at h=1.3 m, thus:

\displaystyle t=\sqrt{\frac{2\cdot 1.3}{9.8}}

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