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steposvetlana
1 month ago
8

A good quarterback can throw a football at 27 m/s (about 60 mph). If we assume that the ball is caught at the same height from w

hich it is thrown, and if we ignore air resistance, what is the maximum range in meters (which is approximately the same as the range in yards) of a pass at this speed? How long is the ball in the air?
Physics
1 answer:
ValentinkaMS [3.4K]1 month ago
8 0

Response:

The ball remained airborne for 3.896 seconds

Explanation:

Given that

g = 9.8 m/s², representing gravitational acceleration,

If the angle of launch is 45°, the horizontal range will be maximized.

Both horizontal and vertical launch velocities are equal, each equating to

v_h  =  v cos θ

v_h  =  27 × cos 45°

         = 19.09 m/s.

The duration to reach maximum height is half of the flight time.

v = u + at   ∵ v = 0 (at maximum height)

19.09 - 9.8 t₁ = 0

t₁ = 1.948 s

The total time in the air equals twice the time to reach maximum height

2 t₁ = 3.896 s

The horizontal distance covered is

D = v × t

D = 3.896×19.09

   = 74.375 m

The ball was in the air for 3.896 seconds

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Two projectiles are launched with the same initial speed from the same location, one at a 30° angle and the other at a 60° angle
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Answer:

Explanation:

Let us denote the launch angle as \theta _1=30^{\circ}.

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H_{max}=\frac{u^2(\sin \theta )^2}{2g}

H_{30}=\frac{u^2(\sin 30)^2}{2g}

H_{30}=\frac{u^2}{8g}

Time of flight is indicated as =\frac{2u\sin \theta }{g}.

t_{30}=\frac{u}{g}

For \theta =60^{\circ}

H_{60}=\frac{u^2(\sin 60)^2}{2g}

H_{60}=\frac{3u^2}{8g}

t_{60}=\frac{2u\sin 60}{g}=\frac{\sqrt{3}u}{g}

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7 0
2 months ago
An elevator is being pulled up from the ground floor to the third floor by a cable. The cable is exerting 4500 newtons of force
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The gravitational force acting on the elevator is 4500N. Explanation: The parameters provided indicate that the force exerted by the elevator is F = 4500 N. The elevator does not accelerate. According to Newton's third law, the force exerted by the cable matches the gravitational force on the elevator, which represents its weight (W), and the elevator's motion stands as follows: F = W + (Mass of elevator × Acceleration of elevator). Therefore, F = W + (Mass of elevator × 0) = W, which leads us to F = 4500 N = W. The net force on the elevator can be expressed as F - W = 0, confirming that the gravitational force on the elevator equals W = 4500N.
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tas watches as his uncle changes a flat tire on a car. his uncle raises the car using a machine called a jack. each time his unc
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The answer is

-Small f and large D.

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-Hence, by applying a small force to the jack, the height at which the car is elevated increases.

Machines are essential for people to amplify their strength; without them, lifting a car would be impossible.

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Answer:

Explanation:

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= 11.34 mA.

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