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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 rise?

Respuesta :

JDcap
[tex]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[/tex]
okpalawalter8

We have that for the Question "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 rise?" it can be said that How high will the rock rise is

S=-11.5m

From the question we are told

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 rise?

Generally the equation for the Motion  is mathematically given as

[tex]v2 - u2 = 2gS \\\\Therefore\\\\S = -(\frac{u^2}{2g}) \\\\S= -(\frac{15^2}{2*9.8})\\\\[/tex]

S=-11.5m

Therefore

How high will the rock rise is

S=-11.5m

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