Basketball on the Moon

Classical Mechanics Level 1

A basketball is dropped from a height of 10 m 10 \text{ m} above the surface of the moon, accelerating downwards at 1.6 m / s 2 1.6 \text { m}/\text{s}^2 . How long does it take to hit the surface, in seconds to the nearest tenth?


The answer is 3.5.

Matt DeCross by Matt DeCross , 27, USA

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6 solutions

Abhiram Rao
May 4, 2016

12 Helpful 0 Interesting 0 Brilliant 2 Confused
Ezra Manatad
Dec 13, 2018

Let s= 10m a=1.6m/s^2 Use S= Vo t + 0.5at^2 10=0.5at^2 10/0.5a = 0.5at^2/0.5a 10/((0.5)(1.6) =t^2 sqrt(12.5)=t t=3.5355 s

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Mariam ElAwady
Oct 27, 2018

d=vi t-1/2 a t^2 10=0-1/5 (-1.6)*t^2 t=3.5s

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Rehman Sazid
Feb 17, 2018

s = ut + 1/2at^2 10 = 0*t + 1/2 * 1.6 *t^2 10 = 0 + 0.8t^2 t^2 = 10/0.8 t = 3.5 sec

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Rohit Iit
Nov 2, 2017
  • We can use the second formula S = ut + 1/2 at^2 From that we get u = 0, then 10 = (0.5)* 1.6 * t^2 10 0.5 ∗ 1.6 \frac{10}{0.5 * 1.6} = t^2 from that we get t = apprx. 3.5 s
0 Helpful 0 Interesting 0 Brilliant 0 Confused
Setu Doshi
Jul 21, 2017

Apply the second eqn. of motion.Take negative value for accelaration and displacement (height).

0 Helpful 0 Interesting 0 Brilliant 0 Confused

U mean the 3rd, right?

Aaditya Singh - 9 months, 4 weeks ago

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