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Calculate the maximum height a ball reaches when thrown straight up at a velocity of 10m/s

Sagot :

Answer:

5.1m

Explanation:

Use the fourth key equation of accelerated motion

vf^2=vi^2+2aΔd

we know vf is = 0 because that is when the ball reaches it's peak (not going up anymore)

0=vi^2+2aΔd

and we can rearrange for Δd, which is what you need

-vi^2=2aΔd

Δd = -vi^2/2a

Now we can plug in the values

Δd = -vi^2/2a

Δd = -(10m/s)^2/2(-9.8)

Δd = -100/-18.6

Δd = 5.10m