Physics, 26.06.2019 23:40 liaholmes8

Abullet is shot horizontally from shoulder height (1.5 m) with an initial speed 200 m/s. (a) how much time elapses before the bullet hits the ground? (b) how far does the bullet travel horizontally?

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Answer from: iisofieii1978

Velocity is the ratio of displacement and time. The time needed by the bullet before it hits the ground is 0.553 sec.

What is Velocity?

Velocity can be defined as the rate of the position of an object with respect to time. It is given by the formula,

$v = \dfrac{dx}{dt}$

Given to us

Height, h = 1.5 m

Initial velocity of the object, v = 200 m/s

Bullet is shot horizontally, θ = 0°

A.) The time needed by the bullet, t can be calculated using the projectile motion equation for time,

$t = [V \times sin(\theta ) + \dfrac{\sqrt{(V\ sin(\theta)^2 + 2 \times g \times h } }{g}$

Substitute the value, we know that the value of sin (0°) = 0

$t = \dfrac{\sqrt{2gh}}{g}\\\\t= \dfrac{\sqrt{2\times 9.81 \times 1.5}}{9.81}\\\\t = 0.553\ sec$

Hence, the time needed by the bullet before it hits the ground is 0.553 sec.

B.) Time needed before the bullet hits the ground,

We know that the motion of the bullet will be projectile, therefore,

$x = v \ cos\theta \times t$

Substitute the value of time,

$x = v\ cos\theta \times t\\\\x = 200\ cos(0^o)\times 0.553\\\\x = 110.6 \ m$

Hence, the distance traveled by the bullet is 110.6 m.

Learn more about Velocity:

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Answer from: mya9859

(a)t=0.553s, (b)x=110.656m

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the bullet has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=200m/s$ is the bullet's initial speed

$\theta=0$ because we are told the bullet is shot horizontally

is the time since the bullet is shot until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=1.5m$ is the initial height of the bullet

y=0 is the final height of the bullet (when it finally hits the ground)

$g=9.8m/s^{2}$ is the acceleration due gravity

Part (a):

Now, for the first part of this problem, the time the bullet elapsed traveling, we will use equation (2) with the conditions given above:

$0=1.5m+200m/s.sin(0) t-\frac{9.8m/s^{2}.t^{2}}{2}$ (3)

$0=1.5m-\frac{9.8m/s^{2}.t^{2}}{2}$ (4)

Finding :

$t=\sqrt{\frac{1.5m(2)}{9.8m/s^{2}}}$ (5)

Then we have the time elapsed before the bullet hits the ground:

t=0.553s (6)

Part (b):

For the second part of this problem, we are asked to find how far does the bullet traveled horizontally. This means we have to use the equation (1) related to the x-component:

$x=V_{o}cos\theta t$ (1)

Substituting the knonw values and the value of found in (6):

x=200m/s.cos(0)(0.553s) (7)

x=200m/s(0.553s) (8)

Finally:

x=110.656m

Answer from: Quest

newton's first law of motion

Answer from: Quest

Yup no bull frogs here man

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