A tennis ball gun launches tennis balls at 18.2 . It\'s pointed straight up and

Question

A tennis ball gun launches tennis balls at 18.2 . It's pointed straight up and launches one ball; 2.0 later, it launches a second ball.

At what time (after the first launch) are the two balls at the same height?


What is that height?


What is the first balls' velocity at that point?


What is the second balls' velocity at that point?

Explanation / Answer

Let the two ball meet after t sec of launch of first ball:- =>By s = ut - 1/2gt^2 =>1st ball:- =>s1 = ut - 1/2gt^2 ----------(i) =>2nd ball:- =>s2 = u(t-2) - 1/2g(t-2)^2 =>s2 = ut - 1/2gt^2 - 2u + 2gt - 2g -----------(ii) =>But s1 = s2 =>ut - 1/2gt^2 = ut - 1/2gt^2 - 2u + 2gt - 2g =>t = (u+g)/g =>t = (17.8+9.8)/9.8 =>t = 2.82 sec

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