2 years ago

A rock is thrown straight up with an initial velocity of 15.0 m/s. Ignore energy lost to air friction. How high will the rock ri

se?

1 answer:

5 0

$h=v_0t- \frac{1}{2} gt^2\\h=15t-\frac{1}{2}(9.8) t^{2}\\ \\ 
v_f=v_0-gt\;\;\;\Rightarrow t= \frac{v_f-v_0}{-g} = \frac{0-15}{-9.8} \approx1.53 \\ \\ 
\Rightarrow h=15(1.53)-4.9(1.53)^2\;\;\;\Rightarrow h=11.48\,m$

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