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postnew [5]
3 years ago
8

Jason and Guy are throwing watermelons straight up from the ground. (They’re making fruit salad.) Jason throws his watermelon at

speed v and it spends time t in the air before smashing into the ground. Not to be outdone, Guy then throws his watermelon at speed 2v. How long is Guy’s watermelon in the air before it smashes back into the ground?
Physics
1 answer:
NISA [10]3 years ago
5 0

Answer:2t

Explanation:

Given

Jason throwing watermelon at speed v and it spend t time in air

time to reach max height

v_f=u_i+at'

0=v-gt'

t'=\frac{v}{g}

and time to reach bottom is also t'

t=2t'

t=2\frac{v}{g}

For guy throws his watermelon at speed 2v

So total time in air will be

t''=2\frac{2v}{g}

t''=4\frac{v}{g}

t''=2t

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At the highest point in its trajectory, the ball's acceleration is zero but its velocity is not zero.

<h3>What's the velocity of the ball at the highest point of the trajectory?</h3>
  • At the highest point, the ball doesn't go more high. So its vertical velocity is zero.
  • However, the ball moves horizontal, so its horizontal component of velocity is non - zero i.e. u×cosθ.
  • u= initial velocity, θ= angle of projection

<h3>What's the acceleration of the ball at the highest point of projectile?</h3>
  • During the whole projectile motion, the earth exerts the gravitational force with a acceleration of gravity along vertical direction.
  • But as there's no acceleration along vertical direction, so the acceleration along vertical direction is zero.

Thus, we can conclude that the acceleration is zero and velocity is non-zero at the highest point projectile motion.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: Player kicks a soccer ball in a high arc toward the opponent's goal. At the highest point in its trajectory

A- neither the ball's velocity nor its acceleration are zero.

B- the ball's acceleration points upward.

C- the ball's acceleration is zero but its velocity is not zero.

D- the ball's velocity points downward.

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brainly.com/question/24216590

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7 0
2 years ago
A radio-controlled car increases its kinetic energy from 4 j to 12 j over a distance of 2 m. what was the average net force on t
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well if we know the meters, then the real question is written as:
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3 0
4 years ago
An object is traveling with a constant velocity of 5 m/s. How far will it have gone after 7 s?
kozerog [31]
<h3>Answer:</h3>

35 meters

<h3>Explanation:</h3>

<u>Data given;</u>

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  • Time taken = 7 s

We are required to calculate how far the object traveled.

  • We need to know that;

Velocity = Displacement ÷ time

  • Therefore;

Displacement = Velocity × time

                       = 5 m/s × 7 s

                      = 35 m

Therefore; the object traveled 35 meters

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3 years ago
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Acceleration is given by:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time interval

Let's apply the formula to the different parts of the problem:

A) -20.5 m/s^2

Let's convert the quantities into SI units first:

u = 19300 km/h \cdot \frac{1000 m/km}{3600 s/h} = 5361.1 m/s

v=1600 km/h  \cdot \frac{1000 m/km}{3600 s/h}  =444.4 m/s

t = 4.0 min = 240 s

So the acceleration is

a=\frac{444.4 m/s-5361.1 m/s}{240 s}=-20.5 m/s^2

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As before, let's convert the quantities into SI units first:

u = 444.4 m/s

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t = 94 s

So the acceleration is

a=\frac{89.2 m/s - 444.4 m/s}{94 s}=-3.8 m/s^2

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For this part we have to use a different formula:

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v = 0 is the final velocity

u = 89.2 m/s is the initial velocity

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d = 75 m is the distance covered

Solving for a, we find

a=\frac{v^2-u^2}{2d}=\frac{0^2-(89.2 m/s)^2}{2(75 m)}=-53.0 m/s^2

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