Question

A body is projected vertically down from the top of the tower. If it travels a distance 30m in one second, the velocity of projection is
1) 34.0 m/s
2) 25.1 m/s
3) 30 m/s
4) zero
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Answer 1

GIVEN:-

• $\rm{Distance\:travelled= 30m}$

• $\rm{time=1s.}$

• $\rm{Acceleration\:due\:to\:Gravity=9.8m/s^{-2}}$

TO FIND:-

• The velocity of Projection i.e Initial Velocity

FORMULAE USED:-

• ${\boxed{\rm{\blue{S=ut+\dfrac{1}{2}\times{a}\times{(t)^2}}}}}$

Now,

Using the equation of motion

$\implies\rm\blue{S=ut+\dfrac{1}{2}\times{a}{(t)^2}}$

$\implies\rm\blue{30m=u(1)+\dfrac{1}{2}\times{9.8}\times{(1)^2}}$

$\implies\rm\red{30=u+\dfrac{1}{\cancel{2}}\times{\cancel{9.8}}\times{1}}$

$\implies\rm\pink{30=u+4.9}$

$\implies\rm\blue{u=30-4.9}$

$\implies\rm\red{u=25.1m/s}$.

Hence, The Option 2 is correct.

MORE TO KNOW

• When Object is thrown vertically upward we take Final Velocity as zero

• When object is falling freely we take initial Velocity as Zero.

• The acceleration due to gravity at the center of the earth is zero

• The Force of Gravitation at the Pole is maximum in Pole and minimum in equator.

Answer 2

Option (2) u = 25.1 m/s .

Given that :-

• Displacement (s) = 30m
• Time (t) = 1sec
• Acceleration due to gravity = 9.8m/s².

To Find :-

• The velocity of projection.

Solution :-

As we know that,

• S = ut + 1/2 at²

=> 30 = u × 1 + 1/2 × 9.8 × (1)²

=> 30 = u + 4.9 × 1

=> 30 = u + 4.9

=> u = 30 - 4.9

=> u = 25.1 m/s .

Hence, the velocity of projection is 25.1 m/s .