A ball is thrown vertically upward it returns back in the hand of the thrower in 60 seconds how high does it go?

Answer 1

Assuming the ball is thrown in a constant gravitational field causing an acceleration of 9.81 m/s/s (similar to Earth's surface),

y(t) = H + Vt - (g/2)(t^2)

Where H is its original height above your chosen origin, V is its initial upward velocity, g is the acceleration due to gravity (9.81 m/s/s) and t is the number of seconds after throwing it. If we set H to 0 (arbitrarily making the ball's initial height 0) and if are interested in a path wherein the ball takes 60 seconds to return to its original height (0), we say:

y(60) = 0 + V(60) - (g/2)(60^2)

The problem states that it "returns back to the hand" at 60 seconds. This implies that at 60 seconds, it is back at its original position (y(60) = 0)

therefore:

y(60) = 0 = 0 + V(60) - (g/2)(60^2)

We can simplify the expression on the RHS above to find V.

The equation for projectile motion is parabolic with respect to time - this means that, by symmetry, the greatest height is achieved at half the time it takes to return. Plug 30 seconds into the above equation, and I get a result of 4414.5 m. That's almost three miles high!