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arsen [322]
2 years ago
10

Physics help please

Physics
1 answer:
zhuklara [117]2 years ago
5 0

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: <u>x-component</u> and <u>y-component.</u> Being their main equations as follows:

<u>x-component: </u>

x=V_{o}cos\theta t   (1)

Where:

x=52 m is the point where the ball strikes ground horizontally

V_{o} is the ball's initial speed

\theta=0 because we are told the ball is thrown horizontally

t is the time since the ball is thrown until it hits the ground

<u>y-component: </u>

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

Where:

y_{o}=120m  is the initial height of the ball

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

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

Knowing this, let's start by finding t from (2):

<u></u>

0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}   (3)

0=y_{o}+\frac{gt^{2}}{2}  

t=\sqrt{\frac{-2 y_{o}}{g}}   (4)

t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}   (5)

t=4.948 s   (6)

Then, we have to substitute (6) in (1):

x=V_{o}cos(0\°) t   (7)

And find V_{o}:

V_{o}=\frac{x}{t}   (8)

V_{o}=\frac{52 m}{4.948 s}   (9)

V_{o}=10.509 m/s   (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity V:

V=V_{o} + gt (11)

V=10.509 m/s + (-9.8 m/s^{2})(4.948 s) (12)

V=-37.981 m/s (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the <u>ball's final speed</u>, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

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So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

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Now, plug in the given values and simplify. This gives,

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