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julia-pushkina [17]
2 years ago
13

A tennis player tosses a tennis ball straight up and then catches it after 2.07 s at the same height as the point of release.

Physics
1 answer:
prisoha [69]2 years ago
3 0

(a) The ball will accelerate at a speed of 9.81 m/s² while it is in flight.

(b) The ball will be moving at a speed of 0 m/sec when it reaches its highest point.

(c) The ball's initial upward velocity is 20.30 m/s.

(d) It will reach a maximum height of 21.0 m.

<h3 /><h3>What is speed?</h3>

Speed is defined as the change in distance with regard to time. A scalar quantity, speed. It is a temporal element, m/sec is its unit.

The information provided in the issue is;

u is the fall's beginning speed in m/sec;

h is the fall's distance

g is the fall's acceleration or 9.81 m/sec²

a. The ball's speed while in flight is equal to the acceleration caused by gravity, which is 9.81 m/s².

b. When the ball reaches its highest point, its velocity will be zero.

c. The ball's starting velocity is;

v=u-gt

0=u-gt

u=gt

u=9.81 m/s² × 2.07 sec

u=20.30 m/s

d. The formula is used to determine the highest height it can reach.

\rm u^2 = - 2gh \\\\ (20.30)^2 = -2 \times (-9.8) \times h \\\\ 412.09  = 19.6 \times h\\\\ h = 412.09/19.6\\\\ h = 21.00 m

As a result, the ball's greatest height, beginning velocity, and acceleration are all 9.81 meters per second, zero meters per second, 20.30 meters per second, and 21 meters accordingly.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

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An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.  

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

\sum F_x = \sum F_y = 0

\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }

  • So; On the L.H.S

\mathbf{mg sin \theta =f_s}

\mathbf{mg sin \theta =\mu_s N}

  • On the R.H.S

N = mg cos θ

Equating both force component together, we have:

\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}

\mathbf{sin \theta =\mu_s \ \ cos \theta}

\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}

From trigonometry rule:

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\mathbf{\mu_s =\tan \theta}}

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