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3 years ago

7

If you throw an object straight up into the air with an initial velocity of 42m/s. What is it’s velocity at the peak of its fl

ight?
A. 42m/s
B. -42m/s
C. 9.8m/s
D. -9.8m/s
E. 0m/s
F. 0m/s2
G. None of these

1 answer:

solniwko [45]3 years ago

7 0

0 m/s. It’s trivial

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A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.0 s after being hit. Then 2.5 s afte

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Answer:

Part a)

$H = 44.1 m$

Part b)

$y = 13.48 m$

Part c)

$d = 8.86 m$

Explanation:

Part a)

As we know that ball will reach at maximum height at

t = 3 s

now we will have

$t = \frac{v sin\theta}{g}$

now we have

$3 = \frac{vsin\theta}{9.8}$

$v sin\theta = 29.4 m/s$

Now maximum height above ground is given as

$H = \frac{v^2sin^2\theta}{2g}$

$H = \frac{29.4^2}{2(9.8)}$

$H = 44.1 m$

Part b)

Height of the fence is given as

$y = (vsin\theta) t - \frac{1}{2}gt^2$

$y = (29.4)(5.5) - \frac{1}{2}(9.8)(5.5^2)$

$y = 13.48 m$

Part c)

As we know that its horizontal distance moved by the ball in 5.5 s is given as

$x = v_x t$

$97.5 = v_x (5.5)$

$v_x = 17.72 m/s$

now total time of flight is given as

$T = 3 + 3 = 6 s$

so range is given as

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4 years ago

When a golf club hits a 0.0459 kg ball at rest, it exerts a 2380 N force for 0.00100 s. What is the speed of the ball afterwards

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Answer:

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Given parameters:

Mass of ball = 0.0459kg

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Unknown:

Speed of the ball afterwards = ?

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To solve this problem, we use Newton's second law of motion:

F = m x $\frac{v - u}{t}$

F is the force

m is the mass

v is the final velocity

u is the initial velocity

t is the time taken

2380 = 0.0459 x $\frac{v- 0}{0.001}$

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3 years ago

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A pilot drops a bomb from a plane flying horizontally with constant velocity. When the bomb hits the ground, the horizontal loca

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Answer: The horizontal location of the plane will BE OVER THE BOMB

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Once the atomic number of an atom is known, the number of electrons can be deduced depending on if the atom is an ion or a neutral one.

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Answer:

Option C is the correct answer.

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