a ball is dropped from a height of 120 meters. it takes 2 seconds for the ball to fall 60 meters, how long before the ball hits the ground?
\(\Huge{\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{orange}{\bigstar}\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{orange}{\bigstar}\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}}\\\color{white}{.}\\\Huge\color{blue}{\mathfrak{~~~~Welcome~to~OpenStudy!~\ddot\smile}}\\\color{white}{.}\\\\\Huge{\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{orange}{\bigstar}\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}\color{orange}{\bigstar}\color{red}{\bigstar}\color{blue}{\bigstar}\color{green}{\bigstar}\color{yellow}{\bigstar}}\)
Since the ball has been droppede, its initial velocity will be 0 m/s. Now using the equation S=ut+1/2at\(^2\) we will find first the value of a 60=0*2+1/2a*2*2 60=2a a=30 m/s\(^2\)
Now using this value of a we can calculate the time taken to travel 120 m by the ball. 120=1/2*30*t\(^2\) 8=t\(^2\) t=2.82 seconds
This question is not very well defined. @Abhisar is correct in one interpretation, but it does assume that the experiment is taking place on another planet where gravity is much greater than earth. I agree that 'dropped' generally means that the initial velocity is 0, however, IF the ball had in initial downward velocity, AND we assume that the acceleration is g (i.e. we are on earth) then (taking g as 10m/s for ease of use...) s=ut+0.5 at^2 60= 2u +5*2^2 allows us to calculate u then 120 = ut +5 t^2 will give t for the total fall. ( Not saying I am correct- just that the question is a little open to interpretation....)
@MrNood since in question they have given that "it takes 2 seconds for the ball to fall 60 meters" this means that the accelaration is greater than g i.e 9.8 m/s
This question is not very well defined. @Abhisar is correct in one interpretation, but it does assume that the experiment is taking place on another planet where gravity is much greater than earth. I agree that 'dropped' generally means that the initial velocity is 0, however, IF the ball had in initial downward velocity, AND we assume that the acceleration is g (i.e. we are on earth) then ( taking g as 10m/s^2) s=ut+0.5 at^2 60= 2u +5*2^2 allows us to calculate u then 120 = ut +5 t^2 will give t for the total fall. ( Not saying I am correct- just that the question is a little open to interpretation....)
@Abhisar As I said above - I do not question your calculation IF the initial velocity is 0 However - the acceleration ON EARTH is g - 9.81m/s^2, so if your assumption is correct then this experiment is NOT taking place on earth. MY assertion is that IFthe ball had an initial downward velocity then it CAN travel 60m in 2 s under the normal acceleration of gravity.
hmm..i agree u ! @MrNood
Join our real-time social learning platform and learn together with your friends!