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Two steel balls of mass 1kg and 2kg and a lead ball of 10kg are released together from the top of tower 30m high. Assuming the path to be in vacuum.(A) The lead ball reaches the ground earlier(B) The 1kg ball steel ball reaches the ground earlier(C) All the balls reach the ground simultaneously(D) The 2kg steel ball reaches the ground earlier.

Last updated date: 20th Jun 2024
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Hint If you notice keenly you will easily find the solution to this question but for sake you can also prove it. For that find the force acting on each of the balls now calculate the acceleration of each ball from that acceleration you can find the time for the balls to reach ground. Assume they are falling without air resistance.

The force due to gravity is given by:
${F_g} = mg$
Where
${F_g}$ is force on ball due to gravity
$m$ is the mass of the ball
$g$ is the acceleration due to gravity
Now force on each ball is:
${F_{steel(1)}} = 1 \times g$
${F_{steel(2)}} = 2 \times g$
${F_{kead}} = 10 \times g$
We know that $F = ma$
So ${F_{steel(1)}} = 1 \times g = 1 \times a$
${F_{steel(2)}} = 2 \times g = 2 \times a$
${F_{kead}} = 10 \times g = = 10 \times a$
We find that $a$ (acceleration) in each case is the same i.e. $g$ therefore, they will reach the ground at the same time.

Hence option C is correct.

Note Pointed to be noted is that, if we consider the case of resistance free falling then a feather and a stone will take the same time to reach the ground if they are left to fall from the same height, in fact their velocity will also be the same. But in real life there are many hindrances which slows down the object like air resistance.