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3 years ago

12

A projectile is launched

1 answer:

Flauer [41]3 years ago

4 0

The vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.

The given parameters;

initial horizontal velocity, vₓ = 16 m/s
initial vertical velocity, $v_y =0$
time interval 1 seconds

The components of the velocity can be horizontal or vertical velocity.

The vertical component of the velocity is affected by acceleration due to gravity while the horizontal component of the velocity is not affected by gravity.

The vertical component of the velocity is calculated as;

$v_y = v_0_y -gt\\\\v_y = 0 - (1\times 9.8)\\\\v_y = -9.8 \ m/s$

The horizontal component of the velocity is constant since it is not affected by gravity.

The horizontal component of the velocity = 16 m/s

Thus, the vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.

Learn more here:brainly.com/question/20349275

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A Neglecting air resistance, a ball projected straight upward so it remains in the air for 10 seconds needs an initial speed of

ololo11 [35]

Answer:

The initial velocity is 50 m/s.

(C) is correct option.

Explanation:

Given that,

Time = 10 sec

For first half,

We need to calculate the height

Using equation of motion

$v^2=u^2+2gh$

$h =\dfrac{v^2}{2g}$ ....(I)

For second half,

We need to calculate the time

Using equation of motion

$h =ut+\dfrac{1}{2}gt_{2}^2$

$h=0+\dfrac{1}{2}gt_{2}^2$

$t_{2}=\sqrt{\dfrac{2h}{g}}$

Put the value of h from equation (I)

$t_{2}=\sqrt{\dfrac{2\times v^2}{g^2}}$

$t_{2}=\dfrac{v}{g}$

According to question,

$t_{1}+t_{2}=10$

$t_{1}=t_{2}$

Put the value of t₁ and t₂

$\dfrac{v}{g}+\dfrac{v}{g}=10$

$\dfrac{2v}{g}=10$

$v=\dfrac{10\times g}{2}$

Here, g = 10

The initial velocity is

$v=\dfrac{10\times10}{2}$

$v=50\ m/s$

Hence, The initial velocity is 50 m/s.

3 0

3 years ago

A wood block, after being given a starting push, slides down a ramp at a constant speed. what is the angle of the ramp above hor

Stells [14]

The solution for the problem is:

Constant speed means Fnet = 0.
Let m = mass of wood block and Θ = angle of ramp; then if µk = 0.35 …

The computation would be:

Fnet = 0 = mg (sin Θ) - (µk) (mg) (cos Θ)
mg (sin Θ) = µk (mg) (cos Θ)
µk = tan Θ
Θ = arctan(µk)

= arctan (0.35)

≈ 19.3°

5 0

3 years ago

A force of 720 Newton stretches a spring 4 meters. A mass of 45 Kilograms is attached to the spring and is initially released fr

poizon [28]

Answer:

x(t) = -3sin2t

Explanation:

Given that

Spring force of, W = 720 N

Extension of the spring, s = 4 m

Attached mass to the spring, m = 45 kg

Velocity of, v = 6 m/s

The proper calculation is attached via the image below.

Final solution is x(t) = -3.sin2t

5 0

3 years ago

g a stone with mass m=1.60 kg IS thrown vertically upward into the air with an initial kinetic energy of 470 J. the drag force a

otez555 [7]

Answer:

Height reached will be 28.35 m

Explanation:

Here we can use the work energy theorem to find the maximum height

As we know by work energy theorem

Work done by gravity + work done by friction = change in kinetic energy

$-mgh - F_f h = 0 - \frac{1}{2}mv_i^2$

now we will have

$-1.60(9.8)(h) - 0.900(h) = - 470$

$-16.58 h = -470$

$h = 28.35 m$

so here the height raised by the stone will be 28.35 m from the ground after projection in upward direction

5 0

3 years ago

mario62 [17]

It's D. If it absorbed it would be turning to steam. I am taking honors chem in high school we are learning this.

7 0

3 years ago

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