An artillery shell is fired from a cliff 20.4 m above a valley witha velocity 25

Question

An artillery shell is fired from a cliff 20.4 m above a valley witha velocity 254 m/s at an angle 37.8 º above the horizontal.How many seconds after being fired does the shell reach its maximumheight?

What is the maximum height of the shell in the previous question(in m) above the valley floor?

An artillery shell is fired from a cliff 28.8 m above a valley witha velocity 234 m/s at an angle 30.8 º above the horizontal.How far (in km) has the shell traveled horizontally when it is atits maximum height?

Explanation / Answer

   v = initial velocity    = release angle    g = acceleration of gravity = 9.807m/s^2    ho = initial height    t = time time of maximum height
t= [2vsin()]/g
maximum height
h = [v2sin2()] / (2g) +ho
distance travelled at maximum height
x = [v2sin(2)] / g
plug the given values, derived using kinematicequations.

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