Question

Worldâs tallest building. suppose that you drop a marble from the top of the burj khalifa building in dubai, which is about 830 m tall. if you ignore air resistance, (a) how long will it take for the marble to hit the ground? (b) how fast will it be moving just before it hits?

Answer 1

a) 13 seconds b) 130 m/s The formula for the distance an object moves while under constant acceleration is d = 1/2AT^2. So let's define d as 830 m, A as 9.8m/s^2, and solve for T 830 m = 1/2 9.8 m/s^2 T^2 830 m = 4.9 m/s^2 T^2 Divide both sides by 4.9 m/s^2 169.3878 s^2 = T^2 Take the square root of both sides 13.01491 s = T Since we only have 2 significant figures, round the result to 13 seconds which is the answer to the first part of the question. To find out how fast the marble is moving, just multiply T and A together 13 s * 9.8 m/s^2 = 127.4 m/s Since we only have 2 significant figures, round the result to 130 m/s.

Answer 2

Answer:

It will take $13.01\rm sec$ for the marble to hit the ground.

It will be moving $127.49\rm m/sec$ fast just before it hits.

Explanation:

Given information:

Initial velocity,$u=0\rm m/s$

acceleration due to gravity is positive as (motion of marble is in same direction of motion)

$a=g=9.8 \rm m/s^2$

Height of Burj khalifa $h=830\rm m$

By using equation of motion,

(a) $u=0,$

$h=ut+\frac{1}{2}\times a t^2=\frac{1}{2}gt^2$

$830\rm m=\frac{1}{2}\times 9.8\rm m/s^2\times t^2$

$t=\sqrt\frac{830\times2}{9.8}=13.01\rm sec$

It will take $13.01\rm sec$ for the marble to hit the ground.

(b)By using,

$u=0$

$v=u+at=gt$

$v=9.8\rm m/s^2\times(13.01\rm s)=127.49\rm m/s$

Hence, It will be moving $127.49\rm m/sec$ fast just before it hits.

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