Question

If a ball throwing vertically up ward reach a maximum height at 50 above the level of projection what will be it time taken to reach higher and the speed of the thrown (take g=10m/s2)?

Answer 1

Recall,

`nu_2 = nu_0^2 + 2aDeltay`,

`nu = nu_0 + at`,

and assume `nu = 0` at the peak of its motion.

Let's derive the initial velocity.

`0 = nu_0^2 - 2 * (9.8m)/s^2 * 50m`

`therefore nu_0 approx (31.3m)/s`

Consider that the maximum height is generally proportional to the initial velocity. If we want a higher maximum height, we need to throw the ball faster. Let's decide I want to throw a ball `100"m"` into the air, I need to throw it,

`0 = nu_0^2 - 2 * (9.8m)/s^2 * 100m`

`therefore nu_0 = (44.3m)/s`

and it would take,

`0 = (44.3m)/s - (9.8m)/s^2 * t`

`therefore t approx 4.52s`

to reach that height.