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In uniformly accelerated motira the foll...

In uniformly accelerated motira the following questions hold :
V=Vo + at
`X=Vot + 1//2 at^2`
When X = displacement , V=Velocity at time, Vo= initial velocity, t=time , and a =acceleration. A ball is projected directly upward at a velocity of 15 m/sec.
What is the distance above the ground after 3 seconds ?

A

1.8 m

B

0.9 m

C

0 m

D

3.6 m.

Text Solution

Verified by Experts

The correct Answer is:
B
Choose the upward direction as negative and downward as positive
Then `V_0`=-15m/sec and `a=9.8 m//sec^2` at t=3 seconds, X=?
X=`V_"ot"+1//2at^2`
X=(1-15)(3)+(1/2)(9.8)`(3)^2`
=-45+44.1
X=-0.9 m above ground .
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Knowledge Check

  • In uniformly accelerated motira the following questions hold : V=Vo + at X=Vot + 1//2 at^2 When X = displacement , V=Velocity at time, Vo= initial velocity, t=time , and a =acceleration. A ball is projected directly upward at a velocity of 15 m/sec. What is Its velocity at that point ?

    A
    14.4 m/sec. downward
    B
    14.4 m/sec. upward
    C
    29.4 m/sec. downward
    D
    29.4 m/sec. upward

  • In uniformly accelerated motira the following questions hold : V=Vo + at X=Vot + 1//2 at^2 When X = displacement , V=Velocity at time, Vo= initial velocity, t=time , and a =acceleration. A ball is projected directly upward at a velocity of 15 m/sec. What is the highest point this ball will reach?

    A
    38.66 m
    B
    11.48 m
    C
    9.80 m
    D
    1.53 m.

  • Displacement x at time t is x=t-t^(3) . If v is the velocity and a the acceleration at time t, then :v =

    A
    `(1)/(2)at^(2)`
    B
    `1+(1)/(2) at`
    C
    `ax`
    D
    `a+(1)/(2)t`

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