Thursday, May 23, 2013

A ball is allowed to fall a height of 15 m. Calculate the velocity with which it strikes the ground. To what height will the ball re-bounced if it...

Hello!


The speed of a falling ball is `g*t`  downwards, where g is the gravity acceleration. The height of a falling ball is


`H(t)=H_0-g(t^2)/(2),`


where `H_0` is the initial height.


A ball strikes the ground when H(t)=0, so


`t_1=sqrt((2H_0)/g),`


and the speed will be `V_1=t_1g=sqrt(2H_0g) approx 17(m/s).` This is the answer to the first part.


The kinetic energy before the strike is `mV_1^2/2,` at the end of the re-bouncing it will be zero. The change...

Hello!


The speed of a falling ball is `g*t`  downwards, where g is the gravity acceleration. The height of a falling ball is


`H(t)=H_0-g(t^2)/(2),`


where `H_0` is the initial height.


A ball strikes the ground when H(t)=0, so


`t_1=sqrt((2H_0)/g),`


and the speed will be `V_1=t_1g=sqrt(2H_0g) approx 17(m/s).` This is the answer to the first part.


The kinetic energy before the strike is `mV_1^2/2,` at the end of the re-bouncing it will be zero. The change of the potential energy will be `mgH_1` , where `H_1` is the maximum height after the re-bouncing. So


`mgH_1=0.4m(V_1)^2/2,` or


`H_1=0.4(V_1)^2/(2g)=0.4H_0=6(m).`


This is the answer for the second part.


That said, the first part could be solved quickier using energy considerations (potential energy becomes kinetic), and the second - considering the potential energy before the fall :)


No comments:

Post a Comment

What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?

Exposition A "decidedly amused" Bobby Gillian leaves the offices of Tolman & Sharp where he is given an envelope containing $1...