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zzz [600]
3 years ago
6

A golfer hits a shot to a green. The ball leaves the club at a speed of 20 m/s at an angle 32° above the horizontal. It rises to

its maximum height and then falls down to the green. What is the speed of the ball at maximum height? Ignore air resistance.
Physics
1 answer:
Daniel [21]3 years ago
4 0

Answer:

16.96 m/s

Explanation:

It is the case of projectile motion in which the projectile fires from the ground and again it hit the ground.

The angle of projection is 32 degree and the velocity of projection is 20 m/s

he velocity at the maximum height is equal to the horizontal component of velocity which always remains constant as there is no acceleration along x axis.

Velocity at the highest point = u Cos 32 = 20 x Cos 32 = 16.96 m/s

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Momentum - mass in motion
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Which type of rock is made of tiny bits of animal shells? A. gabbro B. coquina C. clastic sandstone D. limestone
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Read 2 more answers
Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th
aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

3 0
3 years ago
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densk [106]

Answer:10.4 times of initial velocity

Explanation:

Given

Diameter reduced by 69 %

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suppose d is the initial diameter and d_2 diameter is

d_2=d(1-0.69)

d_2=0.31 d

A_2=\frac{\pi d_2^2}{4}

As flow is constant

Q_1=Q_2

d^2v_0=d_2^2v

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6 0
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satela [25.4K]

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