A shell is fired from a cliff that is 36m above a horizontalplane. The muzzle sp

Question

A shell is fired from a cliff that is 36m above a horizontalplane. The muzzle speed of the shell is 80.0m/s and it is fired atan elevation of 25 degrees above the horizontal.
a) What is the horizontal range of the shell? b) What is the velocity of the shell as it strikes theground? Please show steps, thanks! :) A shell is fired from a cliff that is 36m above a horizontalplane. The muzzle speed of the shell is 80.0m/s and it is fired atan elevation of 25 degrees above the horizontal.
a) What is the horizontal range of the shell? b) What is the velocity of the shell as it strikes theground? Please show steps, thanks! :)

Explanation / Answer

Find the y component of the velocity: -- V(y) = 80Sin V(y) = 80Sin(25) V(y) = 33.8 m/s -- Now, firnd the height when velocity is zero: -- v(f)^2 = v(i)^2 + 2*a*d 0 = 33.8^2 + 2*9.8*d d = 1143 / 19.6 d = 58.3 meters -- Total height used to calculate the velocity is: total height = h(o) + d = 36 m + 58.3 m = 94.3 m -- And the vertical velocity, with conservation of energyis: 1/2mv^2 = mgh v = (2gh) v = 43 m/s -- -- Hope this helps.

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