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Levart [38]
2 years ago
13

A man is running on the straight road with the uniform velocity of 3 metre per second calculate the acceleration for produced hi

m​
Physics
1 answer:
sp2606 [1]2 years ago
6 0

Theoritically

the body moving with uniform velocity has acceleration zero.

Mathmatically,

u=3m/s

v=3m/s (since body is moving with uniform velocity)

a= v-u/t

3-3/t

0/t

0m/s.s

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A basketball player is 4.22 m from
max2010maxim [7]

Answer: The height above the release point is 2.96 meters.

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The acceleration of the ball is the gravitational acceleration in the y axis.

A = (0, -9.8m/s^)

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V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))

for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)

P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)

now, the time at wich the horizontal displacement is 4.22 m will be:

4.22m = 9.20*cos(69°)*t

t = (4.22/ 9.20*cos(69°)) = 1.28s

Now we evaluate the y-position in this time:

h =  -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m

The height above the release point is 2.96 meters.

3 0
3 years ago
Read 2 more answers
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