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2 years ago

A projectile is launched with an initial velocity of (40 m/s), at an angle of (30°) above

Physics

1 answer:

zlopas [31]2 years ago

8 0

Answer:

330.5 m

Explanation:

In this case, the object is launched horizontally at 30° with an initial velocity of 40 m/s .

The maximum height will be calculated as;

$h=\frac{v^2_isin^2\alpha }{2g}$

where ∝ is the angle of launch = 30°

vi= initial launch velocity = 40 m/s

g= 10 m/s²

h= 40²*sin²40° / 2*10

h={1600*0.4132 }/ 20

h= 661.1/2 = 330.5 m

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3 years ago

A woman is standing in the ocean, and she notices that after a wavecrest passes, five more crests pass in a time of 50.0 s. Thed

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Answer:

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(a)

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3 years ago

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3 years ago

Your friend, who is in a field 100 meters away from you, kicks a ball towards you with an initial velocity of 16 m/s. Assuming t

LekaFEV [45]

Answer:

Time, t = 5.355 seconds

Explanation:

Given the following data;

Distance = 100 m

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Deceleration = 1 m/s²

To find the time, we would use the second equation of motion;

But since the ball is decelerating, it's acceleration would be negative.

S = ut + ½at²

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

100 = 16t - 0.5t²

200 = 32t - t²

t² + 32t - 200 = 0

Solving the quadratic equation using the quadratic formula;

The quadratic equation formula is;

$x = \frac {-b \; \pm \sqrt {b^{2} - 4ac}}{2a}$

Substituting into the equation, we have;

$x = \frac {-32 \; \pm \sqrt {32^{2} - 4*1*(-200)}}{2*1}$

$x = \frac {-32\pm \sqrt {1024 - (-800)}}{2}$

$x = \frac {-32 \pm \sqrt {1024 + 800}}{2}$

$x = \frac {-32 \pm \sqrt {1824}}{2}$

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3 years ago

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Read 2 more answers