Question #107167

A canon is launched with an initial velocity of 25m/s and at an angle of 53° to the horizontal on a platform 60m high. Find:

I) the maximum height

II) range of the canon

III) The time for the canon to land

I) the maximum height

II) range of the canon

III) The time for the canon to land

Expert's answer

Given:

"v_0=25\\:\\rm m\/s;"

"\\theta=53^{\\circ};"

"y_0=60\\:\\rm m."

(i) The equations of motion of the canon

"y(t)=y_0+v_{0}\\sin\\theta t-\\frac{gt^2}{2}=60+20t-5t^2;\\\\\nx(t)=x_0+v_{0}\\cos\\theta t=15t."Let's find the maximum value of "y(t)":

(iii)

"y(t)=60+20t-5t^2=0, \\longrightarrow t=6\\:\\rm s."(ii) Range

"R=x(6)=15\\times 6=90\\:\\rm m."
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