Answer to Question #86956 in Mechanics | Relativity for Jonathan

Question

A ball throw vertically upward from groud level hits the ground after 4.0s.calculate the maximum height it reached during the journey.

Answer 1

The ball is thrown upwards and returns in 6 seconds, so it spends half the time, or 2 seconds, going up, and another 2 seconds coming down.

The maximum height h is equal to:

$h(2)=v_0\times t-g\frac{t^2}{2} (1)$

At max height, v=0.

In this case, we can write

$0=v_0-9.8\times 2 (2)$

Using the formula (2) we get: v₀=19.6 m/s

Using the formula (1) we get:

$h(2)=19.6\times t-g\frac{t^2}{2} (3)$

Using the formula (3) we get: h(2)=19.6 m.

Answer:

19.6 m