Math, asked by mimiehaw, 10 months ago

the height, y meters, of an object projected directly upwards from the ground can be modelled by y=20t - 5t^2, where t is the time in seconds after it leaves the ground.

a. find the height of the objects 2 seconds after it leaves the ground.
b. when will the object be 15m above the ground?

Answers

Answered by fitgo72
1

Answer:

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Class 11

>>Physics

>>Motion in a Straight Line

>>Speed and Velocity

>>The position of an object moving along x

Question

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The position of an object moving along x-axis is given by x = a + bt

2

, where a = 8.5 m and b = 2.5 m s

−2

and t is measured in seconds. The average velocity of the object between t = 2 s and t = 4 s is

Medium

Solution

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Correct option is C)

Position is given as x=a+bt

2

=8.5+2.5t

2

Position at t=2 s, x

2

=8.5+2.5(2)

2

=18.5 m

Position at t=4 s, x

1

=8.5+2.5(4)

2

=48.5 m

Displacement S=x

2

−x

1

=48.5−18.5=30 m

Time taken t=4−2=2 s

Average velocity V

avg

=

t

S

=

2

30

=15 m/s

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