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Lesechka [4]
1 year ago
6

If a football quarterback can throw the ball at 20.0 m/s, what is the furthest distance the quarterback could throw the ball neg

lecting air resistance?
Physics
1 answer:
kvasek [131]1 year ago
4 0

The furthest distance the quarterback could throw the ball neglecting air resistance is 81.63 meters.

The situation given above is an example of projectile motion.

The maximum distance up to which a projectile can be thrown is known as its range, which is given by the formula,

Range(R) = 2u^2 sin(2θ)/g

Here u is the maximum velocity of the projectile= 20 m/s

θ is the angle at which the projectile is launched = 45°

g is the acceleration due to gravity = 9.8 m/s^2

Putting the above values in the given equation,

The range comes out to be 81.63 meters.

Hence, the furthest distance the quarterback could throw the ball neglecting air resistance is 81.63 meters.

To know more about "projectile motion", refer to the following link:

brainly.com/question/11422992?referrer=searchResults

#SPJ4

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4 0
2 years ago
1-A train travels 100 km to reach town A in one hour and 15 min. The train stops at station A for 45 minutes. Then it travels 15
kirill [66]

Answer 1) : 62.5 km/hour is the average velocity of the train.

2) The final velocity of the car at the end of 75 m is 14.69 m/s

Explanation:

1) Displacement of the train = 100 km + 150 km = 250 km

Total time train took =1 hour 15 min+ 45 min + 2 hours = 240 min = 4 hours

Average velocity=\frac{Displacement}{time}=\frac{250 km}{4 hour}=62.5 km/h

62.5 km/hour is the average velocity of the train.

2) The acceleration of the car, a= 1.2 m/s^2

Distance covered by the car,s = 75 m

Initial velocity of the car ,v_i = 6 m/s

Final velocity of thre car ,v_f=?

Using third equation of motion:

v_{f}^2=v_{i}^2+2as=(6 m/s)^2+2\times 1.2 m/s\times 75 m=216 m^2/s^2

v_{f}=14.69 m/s

The final velocity of the car at the end of 75 m is 14.69 m/s

8 0
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