Answer to Question #86956 in Mechanics | Relativity for Jonathan

Question #86956
A ball throw vertically upward from groud level hits the ground after 4.0s.calculate the maximum height it reached during the journey.
1
Expert's answer
2019-03-26T06:21:02-0400

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)=v0×tgt22(1)h(2)=v_0\times t-g\frac{t^2}{2} (1)

At max height, v=0.

In this case, we can write

0=v09.8×2(2)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×tgt22(3)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


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