Physics, 28.06.2019 01:30 fatboi82

Aprojectile is launched at ground level with an initial speed of 54.5 m/s at an angle of 35.0° above the horizontal. it strikes a target above the ground 2.80 seconds later. what are the x and y distances from where the projectile was launched to where it lands?

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The x and y distances of projectile from the launching point to the landing point is 125 meters and 48.308 meters respectively.

What is projectile motion?

Projectile motion is the motion of the body, when it is thrown in the air taking the action of gravity on it.

For the distance traveled by the object in a projectile motion, we use the following formula.

$y=u_o+\dfrac{1}{2}gt^2$

Here, (g) is the gravity, ( u_o ) is the initial velocity and (t) is time. The

In the given problem, projectile is launched at ground level with an initial speed of 54.5 m/s.

The initial angle of launching the projectile is 35.0° above the horizontal. The time taken to hit the target after the projectile launched is 2.80 seconds.

The x distance of the projectile is the distance traveled by the object in the horizontal direction. Thus the velocity of the projectile is,

$v_x=54.5\cos (35^o)$

The x distance is the product of horizontal component of the velocity to the time taken by it. Therefore

$x=54.5\cos(35)(2.8)\\x=125\rm m$

The y distance of the projectile is the distance traveled by the object in the vertical direction. Thus the initial velocity of the projectile is,

$u_o=54.5\cos (35^o)$

Put this in the above formula for the y distance, we get,

$y=54.5\sin(35)+\dfrac{1}{2}(-9.81)(2.8)^2\\y=48.308\rm m$

Hence, the x and y distances of projectile from the launching point to the landing point is 125 meters and 48.308 meters respectively.

Learn more about the projectile motion here;

link

Answer from: rafiamajd4537

x=125m, y=48.308m

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which we have two components: x-component and y-component. Being their main equations to find the position as follows:

x-component:

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

Where:

$V_{o}=54.5m/s$ is the projectile's initial speed

$\theta=35\°$ is the angle

t=2.80s is the time since the projectile is launched until it strikes the target

is the final horizontal position of the projectile (the value we want to find)

y-component:

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

Where:

$y_{o}=0$ is the initial height of the projectile (we are told it was launched at ground level)

is the final height of the projectile (the value we want to find)

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

Having this clear, let's begin with x (1):

$x=(54.5m/s)cos(35\°)(2.8s)$ (3)

x=125m (4) This is the horizontal final position of the projectile

For y (2):

$y=0+(54.5m/s)sin(35\°)(2.8s)-\frac{(9.8m/s^{2})(2.8s)^{2}}{2}$ (5)

y=48.308m (6) This is the vertical final position of the projectile

Answer from: Quest

less than one you can't create energy

Answer from: Quest

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Aprojectile is launched at ground level with an initial speed of 54.5 m/s at an angle of 35.0° above...