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2 years ago

The launch speed of a projectile is three times the speed it has at its maximum height.what is the elevation angle at launch?

Physics

1 answer:

Sphinxa [80]2 years ago

6 0

Answer:

Given: a projectile of initial launch velocity(V) and launch angle ∅ and no air resistance. At the maximum height, the projectile would have a zero contribution of speed from the vertical component(Vy) Therefore, if we say Vx=Vcos∅ is the only speed the projectile has at the instant of maximum height then we can replace Vx with 1/5V and write 1/5V=Vcos∅. Solving for the the launch angle ∅, gives Inverse Cos(1/5)=78.5 degrees.

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How much work do you do on a 15 N book in lifting it straight up for a distance of<br> 0.40 meters?

astra-53 [7]

Answer:

Work done, W = 6 J

Explanation:

It is given that,

Force of gravity acting on the book, weight of the book is 15 N

We need to find the work done in lifting the book straight up for a distance of 0.4 meters.

The weight of the book is acting in downward direction and the book is lifted straight up, it means angle between them is 180 degrees. Work done is given by :

$W=Fd\cos180\\\\W=15\times 0.4\times \cos180\\\\W=-6\ J$

So, the magnitude of work done in lifting the book is 6 joules.

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3 years ago

A car moving with an initial speed of 25 m/s slows down to a speed of 5 m/s in 10 seconds Calculate a) the acceleration of the c

stealth61 [152]

Answer :

(a) The acceleration of the car is, $-2m/s^2$

(b) The distance covered by the car is, 150 m

Explanation :

By the 1st equation of motion,

$v=u+at$ ...........(1)

where,

v = final velocity = 5 m/s

u = initial velocity = 25 m/s

t = time = 10 s

a = acceleration of the car = ?

Now put all the given values in the above equation 1, we get:

$5m/s=25m/s+a\times (10s)$

$a=-2m/s^2$

The acceleration of the car is, $-2m/s^2$

By the 2nd equation of motion,

$s=ut+\frac{1}{2}at^2$ ...........(2)

where,

s = distance covered by the car = ?

u = initial velocity = 25 m/s

t = time = 10 s

a = acceleration of the car = $-2m/s^2$

Now put all the given values in the above equation 2, we get:

$s=(25m/s)\times (10s)+\frac{1}{2}\times (-2m/s^2)\times (10s)^2$

By solving the term, we get:

$s=150m$

The distance covered by the car is, 150 m

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3 years ago

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sveticcg [70]

the answer is false

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3 years ago

Read 2 more answers

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-Dominant- [34]

Answer:

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3 years ago

When a guitar string is plucked, the guitar string oscillates as a result of waves moving through the string. The vibrations of

oksian1 [2.3K]

Answer:

Explanation:

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if frequency is constant

wavelength ∝ velocity

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2 years ago